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 = 1e9 + 7;
const int maxn = 220;
struct point {
  long long x, y;
  point operator-(point a) { return (point){x - a.x, y - a.y}; }
  point operator+(point a) { return (point){x + a.x, y + a.y}; }
  long long operator*(point a) { return x * a.y - y * a.x; }
} p[maxn];
long long f[maxn][maxn], S;
bool flag[maxn][maxn];
int n;
bool check(point A, point B, point C, point D) {
  if (min(A.x, B.x) > max(C.x, D.x)) return false;
  if (min(C.x, D.x) > max(A.x, B.x)) return false;
  if (min(A.y, B.y) > max(C.y, D.y)) return false;
  if (min(C.y, D.y) > max(A.y, B.y)) return false;
  B = B - A;
  C = C - A;
  D = D - A;
  if (B * C >= 0 && B * D >= 0) return false;
  if (B * C <= 0 && B * D <= 0) return false;
  B = B + A;
  C = C + A;
  D = D + A;
  A = A - D;
  B = B - D;
  C = C - D;
  if (C * A >= 0 && C * B >= 0) return false;
  if (C * A <= 0 && C * B <= 0) return false;
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d%I64d", &p[i].x, &p[i].y);
  for (int i = 1; i <= n; i++) S += p[i] * p[i % n + 1];
  for (int i = 1; i < n; i++)
    for (int j = i + 2; j <= n; j++) {
      long long tmp = S;
      for (int k = i; k < j; k++) tmp -= p[k] * p[k + 1];
      if (abs(S) == abs(tmp + p[i] * p[j]) + abs(S - tmp + p[j] * p[i]))
        flag[i][j] = 1;
      for (int k = 1; k <= n; k++)
        if (k != i && k != j)
          if ((p[k] - p[i]) * (p[j] - p[i]) == 0 &&
              (p[k].x - p[i].x) * (p[k].x - p[j].x) <= 0 &&
              (p[k].y - p[i].y) * (p[k].y - p[j].y) <= 0)
            flag[i][j] = 0;
      for (int k = 1; k <= n; k++)
        if (check(p[k], p[k % n + 1], p[i], p[j])) flag[i][j] = 0;
    }
  for (int i = 2; i <= n; i++)
    for (int j = 1; j + i - 1 <= n; j++)
      if (i == 2)
        f[j][j + 1] = 1;
      else if (flag[j][j + i - 1]) {
        int l = j, r = j + i - 1;
        for (int k = l + 1; k < r; k++)
          f[l][r] = (f[l][r] + f[l][k] * f[k][r]) % mod;
      }
  printf("%I64d\n", f[1][n]);
  return 0;
}
