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;
int n, dp[205][205], M = 1e9 + 7;
struct point {
  long long x, y;
  struct point operator-(struct point A) {
    struct point t;
    t.x = x - A.x;
    t.y = y - A.y;
    return t;
  };
  long long operator*(struct point A) { return x * A.y - y * A.x; }
} p[205];
long long dfs(int l, int r) {
  if (dp[l][r] != -1) return dp[l][r];
  if (r - l <= 1) return dp[l][r] = 1;
  dp[l][r] = 0;
  for (int i = l + 1; i < r; i++) {
    if ((p[l] - p[r]) * (p[i] - p[r]) > 0) {
      dp[l][r] = (dp[l][r] + dfs(l, i) * dfs(i, r)) % M;
    }
  }
  return dp[l][r];
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i].x >> p[i].y;
  long long temp = 0;
  for (int i = 2; i < n; i++) temp += (p[i - 1] - p[0]) * (p[i] - p[0]);
  if (temp < 0) {
    for (int i = 0; i < n / 2; i++) swap(p[i], p[n - i - 1]);
  }
  memset(dp, -1, sizeof(dp));
  cout << dfs(0, n - 1) << endl;
  return 0;
}
