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>
using namespace std;
struct node {
  long long x, y;
  node operator-(const node& b) const {
    node c;
    c.x = x - b.x, c.y = y - b.y;
    return c;
  }
  long long operator*(const node& b) const { return x * b.y - y * b.x; }
} p[222];
const long long mod = 1e9 + 7ll;
long long dp[222][222];
int n;
long long dfs(int i, int j) {
  if (dp[i][j] != -1) return dp[i][j];
  if (i + 1 == j) return dp[i][j] = 1;
  long long ans = 0;
  for (int k = i + 1; k < j; k++)
    if ((p[k] - p[i]) * (p[j] - p[i]) > 0)
      ans = (ans + dfs(i, k) * dfs(k, j) % mod) % mod;
  return dp[i][j] = ans;
}
int main() {
  while (~scanf("%d", &n)) {
    memset(dp, -1, sizeof(dp));
    for (int i = 0; i < n; i++) scanf("%I64d%I64d", &p[i].x, &p[i].y);
    long long s = 0;
    for (int i = 0; i < n; i++) s += p[i] * p[(i + 1) % n];
    if (s < 0)
      for (int i = 0, j = n - 1; i < j; i++, j--) swap(p[i], p[j]);
    printf("%I64d\n", dfs(0, n - 1));
  }
  return 0;
}
