codeforce 849B

B. Tell Your World

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Connect the countless points with lines, till we reach the faraway yonder.

There are n points on a coordinate plane, the i-th of which being (i, yi).

Determine whether it's possible to draw two parallel and non-overlapping lines, such that every point in the set lies on exactly one of them, and each of them passes through at least one point in the set.

Input

The first line of input contains a positive integer n (3 ≤ n ≤ 1 000) — the number of points.

The second line contains n space-separated integers y1, y2, ..., yn ( - 109 ≤ yi ≤ 109) — the vertical coordinates of each point.

Output

Output "Yes" (without quotes) if it's possible to fulfill the requirements, and "No" otherwise.

You can print each letter in any case (upper or lower).

Examples

input

5
7 5 8 6 9

output

Yes

input

5
-1 -2 0 0 -5

output

No

input

5
5 4 3 2 1

output

No

input

5
1000000000 0 0 0 0

output

Yes

Note

In the first example, there are five points: (1, 7), (2, 5), (3, 8), (4, 6) and (5, 9). It's possible to draw a line that passes through points 1, 3, 5, and another one that passes through points 2, 4 and is parallel to the first one.

In the second example, while it's possible to draw two lines that cover all points, they cannot be made parallel.

In the third example, it's impossible to satisfy both requirements at the same time.

这题看了高手的代码才会。

题意：给出n个点的坐标，问这n个点能否在都在两条平行的直线上。

（注意题面第二行给出了横坐标- - 我都是一直以为没有横坐标，读了好久题才发现）

解题思路：两条平行线的斜率k是一样的，对前三个坐标a[1],a[2],a[3]进行处理，因为要求每个点都在线上，那么这三个点之间的斜率：

k1=a[2]-a[1];

k2=a[3]-a[2];

k3=(a[3]-a[1])/2;

肯定至少有一个是直线的真正斜率（这点可以画个图仔细想想），再根据直线方程y=kx+b，把这三个斜率对所有点套一下，如果只有

两个不相等的b值，则输出Yes，否则输出No。

附代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <set>
 5 using namespace std;
 6 const int M = 1111;
 7 double nu[M];
 8 int n;
 9 bool solve(double k){
10     set<double>s;
11     for(int i=1;i<=n;i++){
12         s.insert(nu[i]-i*k);
13     }
14         return s.size()==2;
15 }
16 int main(){
17     scanf("%d",&n);
18     for(int i=1; i<=n; i++) scanf("%lf",&nu[i]);
19     if(solve(nu[2]-nu[1]) || solve(nu[3]-nu[2]) || solve((nu[3]-nu[1])/2))
20     printf("Yes");
21     else printf("No");
22     return 0;
23 }