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 INF = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3fLL;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int maxn = 200 + 10;
const long long mod = 1e9 + 7;
struct node {
  long long x, y;
  long long operator*(const node& rhs) const { return x * rhs.y - y * rhs.x; }
  node operator-(const node& c) const {
    node u;
    u.x = x - c.x;
    u.y = y - c.y;
    return u;
  }
} p[maxn];
int n;
long long dp[maxn][maxn];
long long solve(int l, int r) {
  if (dp[l][r] != -1) return dp[l][r];
  if (l + 1 == r) return dp[l][r] = 1;
  dp[l][r] = 0;
  for (int k = l + 1; k < r; k++) {
    if ((p[l] - p[r]) * (p[k] - p[r]) > 0) {
      dp[l][r] = (dp[l][r] + solve(l, k) * solve(k, r)) % mod;
    }
  }
  return dp[l][r];
}
int main() {
  n = read();
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < n; i++) p[i].x = read(), p[i].y = read();
  long long tmp = 0;
  p[n] = p[0];
  for (int i = 0; i < n; i++) tmp += p[i] * p[i + 1];
  if (tmp < 0) reverse(p, p + n);
  cout << solve(0, n - 1) << endl;
  return 0;
}
