【41.43%】【codeforces 560C】Gerald's Hexagon

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

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 ≤ ai ≤ 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:

【题目链接】:http://codeforces.com/contest/560/problem/C

【题解】



最后的个数其实可以转换成6边形的面积除边长为1的三角形的面积;



把图补上3个角；

然后用大三角形的面积减去3个小三角形的面积就是六边形的面积了；

大三角形和3个小三角形的都是正三角形;

很好算的;



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

//const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

LL a[7];

LL sqr(LL x)
{
    return x*x;
}

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rep1(i,1,6)
        cin >> a[i];
    cout << sqr(a[1]+a[2]+a[3])-(sqr(a[1])+sqr(a[3])+sqr(a[5]));
    return 0;
}