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 MOD = 1000000007;
typedef struct P {
  int x, y;
  P() {}
  P(int x, int y) : x(x), y(y) {}
} P;
P operator-(const P &a, const P &b) { return P(a.x - b.x, a.y - b.y); }
long long operator^(const P &a, const P &b) {
  return (long long)a.x * b.y - (long long)a.y * b.x;
}
long long darea(const P &a, const P &b, const P &c) {
  return (b - a) ^ (c - a);
}
int side(const P &a, const P &b, const P &c) {
  long long ret = darea(a, b, c);
  return (ret > 0 ? 1 : 0) - (ret < 0 ? 1 : 0);
}
int n;
P p[200];
int ways[200][200];
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d%d", &p[i].x, &p[i].y);
  long long sumdarea = 0;
  for (int i = (2); i < (n); ++i) sumdarea += darea(p[0], p[i - 1], p[i]);
  if (sumdarea < 0) reverse(p, p + n);
  for (int i = (0); i < (n - 1); ++i) ways[i][i + 1] = 1;
  for (int len = (2); len < (n); ++len)
    for (int i = (0); i <= (n - len); ++i) {
      int j = i + len;
      for (int k = (i + 1); k <= (j - 1); ++k) {
        if (side(p[i], p[j], p[k]) >= 0) continue;
        ways[i][j] = (ways[i][j] + (long long)ways[i][k] * ways[k][j]) % MOD;
      }
    }
  printf("%d\n", ways[0][n - 1]);
}
int main() {
  run();
  return 0;
}
