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;
struct point {
  long long x, y;
  friend point operator-(const point &i, const point &j) {
    return (point){i.x - j.x, i.y - j.y};
  }
} p[205];
long long f[205][205], s[205][205];
long long cj(point i, point j) { return i.x * j.y - i.y * j.x; }
inline int cg(int x) {
  x %= n;
  if (x == 0) return n;
  return x;
}
inline long long jdz(long long x) {
  if (x < 0) return 0 - x;
  return x;
}
inline bool ok(int i, int k, int j) {
  long long now = cj(p[j] - p[k], p[i] - p[k]);
  if (now == 0) return 0;
  if ((jdz(now) + jdz(s[i][k]) + jdz(s[k][j]) + jdz(s[j][i])) == jdz(s[1][n]))
    return 1;
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) cin >> p[i].x >> p[i].y;
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; ++j)
      for (int k = i + 1; k < j; ++k) {
        s[i][j] += cj(p[k + 1] - p[k], p[i] - p[k]);
      }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j < i; ++j)
      for (int k = cg(i + 1); k != j; k = cg(k + 1)) {
        s[i][j] += cj(p[cg(k + 1)] - p[k], p[i] - p[k]);
      }
  for (int d = 1; d <= n; ++d)
    for (int i = 1; i <= n; ++i) {
      int j = i + d;
      if (j > n) break;
      if (d == 1) {
        f[i][j] = 1;
        continue;
      }
      for (int k = i + 1; k < j; ++k)
        if (ok(i, k, j)) {
          (f[i][j] += f[i][k] * f[k][j] % 1000000007ll) %= 1000000007ll;
        }
    }
  cout << f[1][n] << endl;
  return 0;
}
