Codeforces Round #431 (Div. 2) B. Tell Your World

题面：

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.

思路：

　　因为一条直线最少过一个点，另一条直线则至少过n-1个点。

　　所以直接枚举那过两个点的直线即可，这样是n*n的做法。

　　但是其实没必要枚举那么多直线，只需考虑前三条直线即可。

　　对于三个点1,2,3

　　判断12在同一直线

　　13在同一直线
　　23在同一直线

　　以上有一个符合情况就可以。

　　但是要特判掉所有点都在同一直线的情况

#include <bits/stdc++.h>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-8;
const double pi=acos(-1.0);
const int K=1e6+7;
const int mod=1e9+7;

int n,vis[K];
struct Point
{
    LL x,y;
}pt[K],pp[K];
LL dc(Point &ta,Point &tb,Point &tc)
{
    return (tb.x-ta.x)*(tc.x-ta.x)+(tb.y-ta.y)*(tc.y-ta.y);
}
LL cc(Point &ta,Point &tb,Point &tc)
{
    return (tb.x-ta.x)*(tc.y-ta.y)-(tc.x-ta.x)*(tb.y-ta.y);
}
bool check(void)
{
    int cnt=0;
    for(int i=3;i<=n;i++)
    if(cc(pt[1],pt[2],pt[i])!=0)
        pp[++cnt]=pt[i];
    for(int i=3;i<=cnt;i++)
    if(cc(pp[1],pp[2],pp[i])!=0)
        return 0;
    Point ta,tb,tc;
    ta.x=pt[2].x-pt[1].x,ta.y=pt[2].y-pt[1].y;
    tb.x=pp[2].x-pp[1].x,tb.y=pp[2].y-pp[1].y;
    tc.x=tc.y=0;
    return cnt<2||cc(tc,ta,tb)==0;
}
int main(void)
{
    int ff=0;cin>>n;
    for(int i=1;i<=n;i++)
        scanf("%lld",&pt[i].y),pt[i].x=i;
    for(int i=3;i<=n&&!ff;i++)
    if(cc(pt[i-2],pt[i-1],pt[i])!=0)
        ff=1;
    if(!ff) {printf("NO
");return 0;}
    if(check())
        {printf("YES
");return 0;}
    swap(pt[1],pt[3]);
    if(check())
        {printf("YES
");return 0;}
    swap(pt[2],pt[3]);
    if(check())
        {printf("YES
");return 0;}
    printf("NO
");
    return 0;
}