ACM 第十五天

计算几何基础

练习题

C - Wasted Time

Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.

Mr. Scrooge's signature can be represented as a polyline A₁A₂... A_n. Scrooge signs like that: first it places a pen at the point A₁, then draws a segment from point A₁ to point A₂, then he draws a segment from point A₂ to point A₃ and so on to point A_n, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.

Scrooge signed exactly k papers throughout his life and all those signatures look the same.

Find the total time Scrooge wasted signing the papers.

Input

The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point A_i — integers x_i and y_i, separated by a space.

All points A_i are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.

Output

Print one real number — the total time Scrooges wastes on
signing the papers in seconds. The absolute or relative error should
not exceed 10^- 6.

Examples

Input

2 1
0 0
10 0

Output

0.200000000

Input

5 10
3 1
-5 6
-2 -1
3 2
10 0

Output

6.032163204

Input

6 10
5 0
4 0
6 0
3 0
7 0
2 0

Output

3.000000000

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include<algorithm>
 #include<math.h>
 #include <bits/stdc++.h>
 using namespace std;
 typedef long double ld;
 typedef long long ll;

 ld ans=;
 int main()
 {
     int n,k;
     ld x[],y[];
     scanf("%d%d",&n,&k);

     for(int i=;i<n;i++)
     {
         cin>>x[i]>>y[i];
     }
     for(int i=;i<n-;i++)
     {
          ans+=sqrt((x[i+]-x[i])*(x[i+]-x[i])+(y[i+]-y[i])*(y[i+]-y[i]));
     }
     ans=ans*k/;
     cout<<fixed<<setprecision()<<ans<<'\n';
     return ;
 }

D - Trace

One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.

Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.

Input

The first line contains the single integer n (1 ≤ n ≤ 100). The second line contains n space-separated integers r_i (1 ≤ r_i ≤ 1000) — the circles' radii. It is guaranteed that all circles are different.

Output

Print the single real number — total area of the part of
the wall that is painted red. The answer is accepted if absolute or
relative error doesn't exceed 10^- 4.

Examples

Input

1
1

Output

3.1415926536

Input

3
1 4 2

Output

40.8407044967

Note

In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals π × 1² = π.

In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (π × 4² - π × 2²) + π × 1² = π × 12 + π = 13π

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include<math.h>

 using namespace std;
 const int maxn = ;
 const double pi=acos(-1.0);
 int a[maxn];
 int main()
 {
     int n;
     double ans=;
     scanf("%d",&n);
     for(int i=; i<=n; i++)
         scanf("%d",&a[i]);
     sort(a+,a+n+);
     int flag=;
     for(int i=n; i>=; i--)
     {
         ans+=(double)pi*a[i]*a[i]*flag;
         flag*=-;
     }
     printf("%.10f\n",ans);
     return ;
 }

H - A Problem about Polyline

There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – ....

We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.

Input

Only one line containing two positive integers a and b (1 ≤ a, b ≤ 10⁹).

Output

Output the only line containing the answer. Your answer
will be considered correct if its relative or absolute error doesn't
exceed 10^- 9. If there is no such x then output  - 1 as the answer.

Examples

Input

3 1

Output

1.000000000000

Input

1 3

Output

-1

Input

4 1

Output

1.250000000000

Note

You can see following graphs for sample 1 and sample 3.

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include<math.h>
 #include <bits/stdc++.h>
 using namespace std;
 typedef long double ld;
 typedef long long ll;
 //const int N=2e5+5;
 ld a,b;
 int main()
 {
     cin>>a>>b;
     //y=x-2pc
     //y=-x+2pc    c=(y+x)/2p
     if(a<b)
         puts("-1");
     else
     {
        ll l=,r=1e9;
         ld res=1e15;
         while(l<=r)
         {
             ll mid=l+r>>;
             ld c=(a+b)/(2.0*mid);
             if(c>=b)
                 l=mid+,res=c;
             else
                 r=mid-;
         }
         l=,r=1e9;
         while(l<=r)
         {
             //y=x-2pc
             ll mid=l+r>>;
             ld c=(a-b)/(2.0*mid);
             if(c>=b)
                 l=mid+,res=min(res,c);
             else
                 r=mid-;
         }
                  cout<<fixed<<setprecision()<<res<<'\n';
     }
     return ;
 }

I - Triangle

Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.

The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.

Input

The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.

Output

Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE
if it is impossible to construct any triangle. Remember that you are to
use three sticks. It is not allowed to break the sticks or use their
partial length.

Examples

Input

4 2 1 3

Output

TRIANGLE

Input

7 2 2 4

Output

SEGMENT

Input

3 5 9 1

Output

IMPOSSIBLE

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include<algorithm>
 #include<math.h>
 #include <bits/stdc++.h>
 using namespace std;
 typedef long double ld;
 typedef long long ll;

 bool tri(int a,int b,int c)
 {
     if(a+b>c && a+c>b && b+c>a)
         return true;
     return false;
 }

 int main()
 {
     int a,b,c,d;
     scanf("%d%d%d%d",&a,&b,&c,&d);
     if(tri(a,b,c)||tri(a,b,d) || tri(a,c,d) || tri(b,c,d))
     {
         printf("TRIANGLE\n");
     }
     else if(a+b==c || a+b==d || a+c==d || a+c==b ||a+d==b || a+d==c ||b+c==d || b+d==c || b+c==a || b+d==a || c+d==a || c+d==b)
     {
         printf("SEGMENT\n");
     }
     else
         printf("IMPOSSIBLE\n");
     return ;
 }

K - Gerald's Hexagon

Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to . Then he measured the length of its sides, and found that each of them is equal to an integer number of centimeters. There the properties of the hexagon ended and Gerald decided to draw on it.

He painted a few lines, parallel to the sides of the hexagon. The lines split the hexagon into regular triangles with sides of 1 centimeter. Now Gerald wonders how many triangles he has got. But there were so many of them that Gerald lost the track of his counting. Help the boy count the triangles.

Input

The first and the single line of the input contains 6 space-separated integers a₁, a₂, a₃, a₄, a₅ and a₆ (1 ≤ a_i ≤ 1000)
— the lengths of the sides of the hexagons in centimeters in the
clockwise order. It is guaranteed that the hexagon with the indicated
properties and the exactly such sides exists.

Output

Print a single integer — the number of triangles with the sides of one 1 centimeter, into which the hexagon is split.

Examples

Input

1 1 1 1 1 1

Output

6

Input

1 2 1 2 1 2

Output

13

Note

This is what Gerald's hexagon looks like in the first sample:

And that's what it looks like in the second sample:

 #include<stdio.h>
 int s[];
 using namespace std;
 int main()
 {
     for(int i=;i<=;i++)
         scanf("%d",&s[i]);
     int a=(s[]+s[]+s[]);
     printf("%d\n",a*a-s[]*s[]-s[]*s[]-s[]*s[]);
   return ;
 }

M - Gleb And Pizza

Gleb ordered pizza home. When the courier delivered the pizza, he was very upset, because several pieces of sausage lay on the crust, and he does not really like the crust.

The pizza is a circle of radius r and center at the origin. Pizza consists of the main part — circle of radius r - d with center at the origin, and crust around the main part of the width d. Pieces of sausage are also circles. The radius of the i -th piece of the sausage is r_i, and the center is given as a pair (x_i, y_i).

Gleb asks you to help determine the number of pieces of sausage caught on the crust. A piece of sausage got on the crust, if it completely lies on the crust.

Input

First string contains two integer numbers r and d (0 ≤ d < r ≤ 500) — the radius of pizza and the width of crust.

Next line contains one integer number n — the number of pieces of sausage (1 ≤ n ≤ 10⁵).

Each of next n lines contains three integer numbers x_i, y_i and r_i ( - 500 ≤ x_i, y_i ≤ 500, 0 ≤ r_i ≤ 500), where x_i and y_i are coordinates of the center of i-th peace of sausage, r_i — radius of i-th peace of sausage.

Output

Output the number of pieces of sausage that lay on the crust.

Examples

Input

8 4
7
7 8 1
-7 3 2
0 2 1
0 -2 2
-3 -3 1
0 6 2
5 3 1

Output

2

Input

10 8
4
0 0 9
0 0 10
1 0 1
1 0 2

Output

0

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include<math.h>

 using namespace std;
 const int maxn = ;
 const double pi=acos(-1.0);

 int main()
 {
     int R,d;
     scanf("%d%d",&R,&d);
     int n;
     scanf("%d",&n);
     int ans=;
     int x,y,r;
     while(n--)
     {
         scanf("%d%d%d",&x,&y,&r);
         double distance=sqrt(x*x+y*y);
         double xiao=distance-r;
         double da=distance+r;
         if(xiao>=R-d && da<=R)
         {
             ans++;
         }

     }
     printf("%d\n",ans);
     return ;
 }

Note

Below is a picture explaining the first example. Circles of green color denote pieces of sausage lying on the crust.