This time our child has a simple polygon. He has to find the number of ways to split the polygon into non-degenerate triangles, each way must satisfy the following requirements:

  * each vertex of each triangle is one of the polygon vertex; 
  * each side of the polygon must be the side of exactly one triangle; 
  * the area of intersection of every two triangles equals to zero, and the sum of all areas of triangles equals to the area of the polygon; 
  * each triangle must be completely inside the polygon; 
  * each side of each triangle must contain exactly two vertices of the polygon. 



The picture below depicts an example of a correct splitting.

<image>

Please, help the child. Calculate the described number of ways modulo 1000000007 (109 + 7) for him.

Input

The first line contains one integer n (3 ≤ n ≤ 200) — the number of vertices of the polygon. Then follow n lines, each line containing two integers. The i-th line contains xi, yi (|xi|, |yi| ≤ 107) — the i-th vertex of the polygon in clockwise or counterclockwise order.

It's guaranteed that the polygon is simple.

Output

Output the number of ways modulo 1000000007 (109 + 7).

Examples

Input

4
0 0
0 1
1 1
1 0


Output

2


Input

4
0 0
1 0
0 1
-1 0


Output

1


Input

5
0 0
1 0
1 1
0 1
-2 -1


Output

3

Note

In the first sample, there are two possible splittings:

<image>

In the second sample, there are only one possible splitting:

<image>
#include <bits/stdc++.h>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
const int inf = 0x3f3f3f3f;
const long long INF = LONG_LONG_MAX;
const int N = 205;
struct node {
  long long x, y;
  node() {}
  node(long long x, long long y) : x(x), y(y) {}
  node operator-(const node &rhs) { return node(x - rhs.x, y - rhs.y); }
} p[N];
long long det(node a, node b) { return a.x * b.y - a.y * b.x; }
long long dp[N][N];
const long long mod = 1e9 + 7;
int main() {
  int n;
  long long sum = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld%lld", &p[i].x, &p[i].y);
  for (int i = 2; i <= n; i++) sum += det(p[i], p[i + 1]);
  sum += det(p[n], p[1]);
  if (sum > 0) reverse(p + 1, p + n + 1);
  for (int i = 1; i < n; i++) dp[i][i + 1] = 1;
  for (int len = 3; len <= n; len++) {
    for (int i = 1; i + len - 1 <= n; i++) {
      int j = i + len - 1;
      for (int k = i + 1; k <= j - 1; k++) {
        if (det(p[j] - p[i], p[k] - p[i]) > 0)
          dp[i][j] += dp[i][k] * dp[k][j], dp[i][j] %= mod;
      }
    }
  }
  printf("%lld\n", dp[1][n]);
  return 0;
}
