B

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, thei-th of which being (i, y_i).

Determine whether it's possible to draw two parallel and non-overlapping lines, such that every point in the set lies onexactly one of them, and each of them passes throughat 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 integersy₁, y₂, ..., y_n ( - 10⁹ ≤ y_i ≤ 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

57 5 8 6 9

Output

Yes

Input

5-1 -2 0 0 -5

Output

No

Input

55 4 3 2 1

Output

No

Input

51000000000 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 points1, 3, 5, and another one that passes through points2, 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.

思路：直接枚举前三个点构成的斜率，其中至少有一个是标准的斜率，这样的话分别以这三个斜率作为标准斜率，判断是否构成两条线就可以了。

#include<bits/stdc++.h>using namespace std;int s[3000],n;bool judge(double k){    int p=-1;    for(int i=2; i<=n; i++)    {        if(s[i]-s[1]==(i-1)*k)            continue;        if(p==-1)            p=i;        else if(s[i]-s[p]!=(i-p)*k)            return 0;    }    return p!=-1;}int main(){    while(scanf("%d",&n)!=EOF)    {        for(int i=1; i<=n; i++)            scanf("%d",&s[i]);        if(judge(s[2]-s[1])||judge((s[3]-s[1])*1.0/2)||judge(s[3]-s[2]))            printf("Yes\n");        else            printf("No\n");    }}