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;
const int MAXN = 2e2 + 10;
const long long mod = 1e9 + 7;
long long xi[MAXN];
long long yi[MAXN];
long long dp[MAXN][MAXN];
int lr[MAXN][MAXN][MAXN];
int hefa;
long long check(int i, int j, int k) {
  long long ax = xi[j] - xi[i];
  long long ay = yi[j] - yi[i];
  long long bx = xi[k] - xi[i];
  long long by = yi[k] - yi[i];
  long long sinz = ax * by - bx * ay;
  if (sinz > 0) sinz = 1;
  if (sinz < 0) sinz = -1;
  return sinz;
}
long long cnt(int i, int j) {
  long long ax = xi[i];
  long long ay = yi[i];
  long long bx = xi[j];
  long long by = yi[j];
  long long sinz = ax * by - bx * ay;
  return sinz;
}
long long dfs(int l, int r) {
  if (dp[l][r] >= 0) return dp[l][r];
  if (l == r)
    return dp[l][r] = 0;
  else if (l + 1 == r)
    return dp[l][r] = 1;
  long long res = 0;
  for (int i = l + 1; i < r; ++i) {
    if (!lr[l][r][i]) continue;
    res = (res + dfs(l, i) % mod * dfs(i, r) % mod) % mod;
  }
  return dp[l][r] = res;
}
int main() {
  while (cin >> n) {
    memset(dp, -1, sizeof dp);
    memset(lr, 0, sizeof lr);
    long long sum = 0;
    for (int i = 1; i <= n; ++i) {
      cin >> xi[i] >> yi[i];
      if (i != 1) {
        sum += cnt(i - 1, i);
      }
    }
    sum += cnt(n, 1);
    hefa = sum > 0 ? 1 : -1;
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= n; ++j) {
        if (i == j) continue;
        for (int k = 1; k <= n; ++k) {
          if (k == j || k == i) continue;
          lr[i][j][k] = (check(i, k, j) == hefa) ? 1 : 0;
        }
      }
    }
    cout << dfs(1, n) << endl;
  }
  return 0;
}
