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 N = 201, M = 1e9 + 7;
struct point {
  long long x, y;
  point(long long a = 0, long long b = 0) {
    x = a;
    y = b;
  }
  void sc() { scanf("%lld%lld", &x, &y); }
  void print() { printf("(%lld,%lld)\n", x, y); }
  double ang() { return atan2(y, x); }
  double angdeg() {
    double ret = 180 * atan2(y, x) / acos(-1);
    if (ret < 0) ret += 360;
    return ret;
  }
  point normal() { return point(x / len(), y / len()); }
  double len() { return sqrt(y * y + x * x); }
  point operator+(const point &a) { return point(x + a.x, y + a.y); }
  point operator-(const point &a) { return point(x - a.x, y - a.y); }
  point operator-() { return point(-x, -y); }
  long long operator*(const point &a) { return a.x * x + a.y * y; }
  long long operator^(const point &a) { return x * a.y - a.x * y; }
  point operator*(long long a) { return point(x * a, y * a); }
  point operator/(long long a) { return point(x / a, y / a); }
  bool operator<(point b) {
    if (x != b.x) return x < b.x;
    return y < b.y;
  }
  long long dis(point a) {
    return sqrt((x - a.x) * (x - a.x) + (y - a.y) * (y - a.y));
  }
  point mid(point a) { return point((a.x + x) / 2, (a.y + y) / 2); }
} a[N];
int n, dp[N][N];
void add(int &a, int b) {
  a += b;
  if (a >= M) a -= M;
}
long long area;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    a[i].sc();
    dp[i][i] = dp[i][i + 1] = 1;
  }
  for (int i = 0; i < n; i++) area += a[i] ^ a[(i + 1) % n];
  if (area < 0) reverse(a, a + n);
  for (int len = 2; len < n; len++)
    for (int i = 0; i + len < n; i++) {
      int j = i + len;
      for (int k = i + 1; k < j; k++)
        if (((a[k] - a[i]) ^ (a[i] - a[j])) < 0)
          add(dp[i][j], dp[i][k] * 1LL * dp[k][j] % M);
    }
  printf("%d\n", dp[0][n - 1]);
}
