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 MO = 1000000007;
long long dp[300][300];
int n;
struct Point {
  long long x, y;
  Point() {}
  Point(long long x0, long long y0) : x(x0), y(y0) {}
} P[300];
Point operator-(Point p1, Point p2) { return Point(p1.x - p2.x, p1.y - p2.y); }
long long operator*(Point p1, Point p2) { return p1.x * p2.y - p1.y * p2.x; }
long long work(int x, int y) {
  if (dp[x][y] != -1) return dp[x][y];
  long long ans(0);
  if (y - x == 1) return 1;
  for (int i = x + 1; i <= y - 1; i++) {
    if ((P[x] - P[y]) * (P[i] - P[y]) > 0) {
      ans = (ans + work(x, i) * work(i, y)) % MO;
    }
  }
  return dp[x][y] = ans;
}
int main() {
  scanf("%d", &n);
  memset(dp, -1, sizeof(dp));
  for (int i = 1; i <= n; i++) {
    scanf("%I64d%I64d", &P[i].x, &P[i].y);
  }
  long long sum(0);
  P[n + 1] = P[1];
  for (int i = 1; i <= n; i++) sum += P[i] * P[i + 1];
  if (sum < 0) {
    for (int i = 1; i <= int(n / 2); i++) swap(P[i], P[n - i + 1]);
  }
  printf("%I64d", work(1, n));
}
