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>
long long mod = 1e9 + 7;
using namespace std;
long long dp[300][300], dq, ans;
int n;
struct Node {
  long long x, y;
} p[500];
long long chaji(Node x, Node y) { return x.x * y.y - x.y * y.x; }
long long solve(int l, int r) {
  if (dp[l][r] != -1) return dp[l][r];
  long long dq = 0;
  if (r == l + 1) {
    dp[l][r] = 1;
    return 1;
  }
  for (int i = l + 1; i < r; ++i) {
    Node dq1, dq2;
    dq1.x = p[l].x - p[r].x;
    dq1.y = p[l].y - p[r].y;
    dq2.x = p[i].x - p[r].x;
    dq2.y = p[i].y - p[r].y;
    long long now = chaji(dq1, dq2);
    if (now > 0) dq = (dq + solve(l, i) * solve(i, r)) % mod;
  }
  dp[l][r] = dq;
  return dq;
}
int main() {
  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 dq = 0;
  p[n] = p[0];
  for (int i = 0; i < n; ++i) dq += chaji(p[i], p[i + 1]);
  if (dq < 0) reverse(p, p + n);
  ans = solve(0, n - 1);
  printf("%I64d\n", ans);
  return 0;
}
