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 long long MOD = 1000000007LL;
const int MAXN = 256;
long long px[MAXN], py[MAXN];
bool good[MAXN][MAXN];
long long dp[MAXN][MAXN];
int N;
long long cross(long long x1, long long y1, long long x2, long long y2) {
  return x1 * y2 - x2 * y1;
}
long long dot(long long x1, long long y1, long long x2, long long y2) {
  return x1 * x2 + y1 * y2;
}
int sgn(long long x) {
  if (x < 0)
    return -1;
  else if (x > 0)
    return 1;
  else
    return 0;
}
bool isonline(int a, int b, int c) {
  long long x1 = px[c] - px[a], y1 = py[c] - py[a];
  long long x2 = px[b] - px[a], y2 = py[b] - py[a];
  return cross(x1, y1, x2, y2) == 0 && 0 <= dot(x1, y1, x2, y2) &&
         dot(x1, y1, x2, y2) <= dot(x2, y2, x2, y2);
}
bool ssc(int a, int b, int c, int d) {
  long long x21 = px[b] - px[a], y21 = py[b] - py[a];
  long long x31 = px[c] - px[a], y31 = py[c] - py[a];
  long long x41 = px[d] - px[a], y41 = py[d] - py[a];
  return (sgn(cross(x21, y21, x31, y31)) * sgn(cross(x21, y21, x41, y41)) < 0);
}
bool crossline(int a, int b, int c, int d) {
  return ssc(a, b, c, d) && ssc(c, d, a, b);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> px[i] >> py[i];
  }
  long long area = 0;
  for (int i = 0; i < N; i++) {
    int j = (i + 1) % N;
    area += cross(px[i], py[i], px[j], py[j]);
  }
  if (area < 0) {
    reverse(px, px + N);
    reverse(py, py + N);
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (i == j) continue;
      good[i][j] = true;
      for (int k = 0; k < N; k++) {
        if (k != i && k != j && isonline(i, j, k)) good[i][j] = false;
        if (crossline(i, j, k, (k + 1) % N)) good[i][j] = false;
      }
    }
  }
  for (int i = 0; i < N - 1; i++) dp[i][i + 1] = 1;
  for (int t = 2; t < N; t++) {
    for (int i = 0; i + t < N; i++) {
      int j = i + t;
      for (int k = i + 1; k <= j - 1; k++) {
        if (!good[i][j] || !good[j][k] || !good[k][i]) continue;
        if (cross(px[k] - px[i], py[k] - py[i], px[j] - px[i], py[j] - py[i]) <=
            0)
          continue;
        dp[i][j] += dp[i][k] * dp[k][j];
        dp[i][j] %= MOD;
      }
    }
  }
  long long ans = dp[0][N - 1];
  cout << ans << endl;
  return 0;
}
