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;
long long dp[205][205];
struct Point {
  int x, y;
  Point() {}
  Point(int x, int y) : x(x), y(y) {}
  inline void read() { scanf("%d%d", &x, &y); }
} vertex[205];
Point operator-(const Point &a, const Point &b) {
  return Point(a.x - b.x, a.y - b.y);
}
long long operator*(const Point &a, const Point &b) {
  return (long long)a.x * b.y - (long long)a.y * b.x;
}
long long DP(int l, int r) {
  long long res = 0LL;
  if (dp[l][r] != -1) return dp[l][r];
  if (l + 1 == r) return dp[l][r] = 1LL;
  for (int i = l + 1; i < r; ++i) {
    if ((vertex[r] - vertex[l]) * (vertex[i] - vertex[l]) > 0) {
      res = (res + DP(l, i) * DP(i, r)) % 1000000007;
    }
  }
  return dp[l][r] = res;
}
bool isclockwise() {
  long long area = 0;
  for (int i = 1; i < n; i++) {
    area += vertex[i] * vertex[i + 1];
  }
  area += vertex[n] * vertex[1];
  return area < 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) vertex[i].read();
  memset(dp, -1, sizeof(dp));
  if (!isclockwise()) {
    reverse(vertex + 1, vertex + 1 + n);
  }
  printf("%I64d\n", DP(1, n));
  return 0;
}
