B. Tell Your World (+ geometric mathematical thinking)

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

Copy

5
7 5 8 6 9

output

Copy

Yes

input

Copy

5
-1 -2 0 0 -5

output

Copy

No

input

Copy

5
5 4 3 2 1

output

Copy

No

input

Copy

5
1000000000 0 0 0 0

output

Copy

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.

Algorithm: Geometric mathematical thinking +

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

typedef long long ll;

#define INF 0x3f3f3f3f
const int maxn = 1e5+7;

ll a[maxn];
int n;

int solve(double k) {
    int pos = -1;
    for(int i = 2; i <= n; i++) {
        if(a[i] - a[1] == (i - 1) *K) {
             Continue ; 
        } 
        IF (POS == - . 1 ) { 
            POS = I;         // determine a new point 
        } the else  IF ! (A [I] - A [POS] = (I - POS) * K) {
             return  0 ; 
        } 
    } 
    return ! = POS - . 1 ;    // determines whether all points on a straight line 
} 

int main () {
     the while (~ Scanf ( " % D " &, n-)) {
         for ( int I = 1; I <= n-; I ++ ) { 
            CIN >> A [I]; 
        } 
        // to determine three points three lines, the following three conditions 
        Double K1 = A [ 2 ] - A [ . 1 ];        
         Double K2 = 1.0 * (A [ . 3 ] - A [ . 1 ]) / 2 ;
         Double K3 = A [ . 3 ] - A [ 2 ];
         IF (Solve (K1) || Solve (K2) || Solve (K3)) { 
            the printf ( " Yes \ n- " ); 
        } the else { 
            the printf ( "No\n");
        }
    }
    return 0;
}

B. Tell Your World (+ geometric mathematical thinking)

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