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 = 1 << 28;
const int mod = (int)1e9 + 7;
const double eps = 1e-8;
int n, m, k;
struct node {
  double x, y;
  node operator-(node b) {
    node tmp;
    tmp.x = x - b.x;
    tmp.y = y - b.y;
    return tmp;
  }
} p[210];
double crs(node a, node b) { return a.x * b.y - b.x * a.y; }
long long dp[210][210];
long long dfs(int l, int r) {
  if (dp[l][r] != -1) return dp[l][r];
  if (r - l == 1) return dp[l][r] = 1;
  long long &sum = dp[l][r];
  sum = 0;
  for (int i = l + 1; i < r; i++) {
    if (crs(p[i] - p[l], p[r] - p[i]) > 0) {
      sum = (sum + dfs(l, i) * dfs(i, r) % mod) % mod;
    }
  }
  return dp[l][r];
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    scanf("%lf%lf", &p[i].x, &p[i].y);
  }
  double sum = 0;
  for (int i = 1; i < n - 1; i++) {
    sum += crs(p[i] - p[0], p[i + 1] - p[0]);
  }
  if (sum < 0) reverse(p, p + n);
  memset(dp, -1, sizeof dp);
  cout << dfs(0, n - 1) << endl;
  return 0;
}
