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;
const int N = 205;
const int P = 1000000007;
int n;
long long dp[N][N];
long long x[N], y[N];
long long cross(int k, int i, int j) {
  return (x[i] - x[k]) * (y[j] - y[k]) - (y[i] - y[k]) * (x[j] - x[k]);
}
int solve(int k, int l, int r) {
  long long ans = cross(k, l, r);
  if (ans > 0) return 1;
  if (ans < 0) return -1;
  return 0;
}
bool Judge(int l, int r, int i, int j) {
  int A = solve(l, i, j);
  int B = solve(r, i, j);
  int C = solve(i, l, r);
  int D = solve(j, l, r);
  if (A == 0 && B == 0 && C == 0 && D == 0) return 0;
  return A * B <= 0 && C * D <= 0;
}
bool Judge(int l, int r) {
  long long sum = 0;
  for (int i = l; i <= r - 1; i++) sum += cross(l, i, i + 1);
  if (sum <= 0) return 0;
  sum = 0;
  for (int i = r; i != l; i = i % n + 1) sum += cross(l, i, i % n + 1);
  if (sum <= 0) return 0;
  for (int i = 1; i <= n; i++) {
    int j = i % n + 1;
    if (i == l || i == r) continue;
    if (j == l || j == r) continue;
    if (Judge(l, r, i, j)) return 0;
  }
  return 1;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> x[i] >> y[i];
  long long sum = 0;
  for (int i = 2; i <= n - 1; i++) sum += cross(1, i, i + 1);
  if (sum < 0) {
    reverse(x + 1, x + n + 1);
    reverse(y + 1, y + n + 1);
  }
  for (int i = 1; i <= n - 1; i++) dp[i][i + 1] = 1;
  for (int d = 2; d <= n - 2; d++)
    for (int i = 1; i <= n - d; i++)
      if (Judge(i, i + d)) {
        int j = i + d;
        for (int k = i + 1; k <= j - 1; k++)
          (dp[i][j] += dp[i][k] * dp[k][j]) %= P;
      }
  long long ans = 0;
  for (int k = 2; k <= n - 1; k++) (ans += dp[1][k] * dp[k][n]) %= P;
  cout << ans << endl;
  return 0;
}
