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 MOD = 1000000007;
using Point = pair<int, int>;
int n;
Point points[N];
long long area[N][N];
long long cross(Point p, Point q, Point r) {
  return 1LL * (q.first - p.first) * (r.second - q.second) -
         1LL * (q.second - p.second) * (r.first - q.first);
}
int dp[N][N];
int f(int l, int r) {
  if (r - l == 1) return 1;
  if (~dp[l][r]) return dp[l][r];
  int ret = 0;
  for (int k = l + 1; k < r; ++k) {
    long long triangle = abs(cross(points[l], points[k], points[r]));
    if (triangle == 0) continue;
    if (triangle + area[l][k] + area[k][r] + area[r][l] > area[0][n - 1])
      continue;
    ret = (1LL * f(l, k) * f(k, r) + ret) % MOD;
  }
  return dp[l][r] = ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d %d", &points[i].first, &points[i].second);
  }
  for (int i = 0; i < n; ++i) {
    area[i][i] = 0;
    for (int j = 1; j < n; ++j) {
      area[i][(i + j) % n] =
          area[i][(i + j - 1) % n] +
          cross(points[i], points[(i + j - 1) % n], points[(i + j) % n]);
    }
    for (int j = 0; j < n; ++j) area[i][j] = abs(area[i][j]);
  }
  memset(dp, -1, sizeof dp);
  printf("%d\n", f(0, n - 1));
  return 0;
}
