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 inf = 1e9;
const long long INF = 1e18;
const double eps = 1e-10;
const int maxn = 220;
int dp[maxn][maxn];
int dcmp(double x) { return x < -eps ? -1 : x > eps; }
typedef struct P {
  double x, y;
  P(double x = 0, double y = 0) : x(x), y(y) {}
  P operator+(P r) { return P(x + r.x, y + r.y); }
  P operator-(P r) { return P(x - r.x, y - r.y); }
  P operator*(double r) { return P(x * r, y * r); }
  P operator/(double r) { return P(x / r, y / r); }
  bool operator<(const P &r) const { return x < r.x || (x == r.x && y < r.y); }
  bool operator==(const P &r) const {
    return dcmp(x - r.x) == 0 && dcmp(y - r.y) == 0;
  }
  bool upper() { return y > 0 || (y == 0 && x > 0); }
  void read() { scanf("%lf %lf", &x, &y); }
  void print() { printf("%.2f %.2f\n", x, y); }
} V;
double Cross(V A, V B) { return A.x * B.y - A.y * B.x; }
P a[maxn];
int n;
const int mod = 1e9 + 7;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) a[i].read();
  double s = Cross(a[1], a[n]);
  for (int i = 2; i <= n; i++) s += Cross(a[i], a[i - 1]);
  if (s < 0) {
    for (int i = 1; i <= n / 2; i++) swap(a[i], a[n - i + 1]);
  }
  for (int i = 1; i < n; i++) dp[i][i + 1] = 1;
  for (int len = 3; len <= n; len++) {
    for (int l = 1; l + len - 1 <= n; l++) {
      int r = l + len - 1;
      for (int k = l; k <= r; k++) {
        if (Cross(a[r] - a[l], a[k] - a[l]) > 0)
          dp[l][r] = (dp[l][r] + 1ll * dp[l][k] * dp[k][r]) % mod;
      }
    }
  }
  printf("%d\n", dp[1][n]);
  return 0;
}
