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, x[210], y[210], ans, dp[210][210];
long long S[210][210];
bool in[210][210][210];
int gogo(int x) {
  if (x % n == 0) return n;
  return x % n;
}
long long cp(int a, int b, int c) {
  return 1LL * (x[a] - x[c]) * (y[b] - y[c]) -
         1LL * (y[a] - y[c]) * (x[b] - x[c]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d%d", &x[i], &y[i]);
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      for (int p = i + 1; p < j; p++) S[i][j] += cp(p, p + 1, i);
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j < i; j++) {
      for (int p = gogo(i + 1); p != j; p = gogo(p + 1))
        S[i][j] += cp(gogo(p), gogo(p + 1), i);
    }
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      for (int p = j + 1; p <= n; p++)
        if (abs(cp(j, p, i)) + abs(S[i][j]) + abs(S[j][p]) + abs(S[p][i]) ==
                abs(S[1][n]) &&
            cp(j, p, i)) {
          in[i][j][p] = true;
        }
  for (int i = 1; i <= n; i++) dp[i][i + 1] = 1;
  for (int i = n; i; i--)
    for (int j = i + 1; j <= n; j++)
      for (int p = i + 1; p < j; p++)
        if (in[i][p][j])
          dp[i][j] = (dp[i][j] + 1LL * dp[i][p] * dp[p][j]) % 1000000007;
  ans = dp[1][n];
  printf("%d\n", ans);
}
