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;
struct Point {
  long long x, y;
  Point(long long _x = 0, long long _y = 0) : x(_x), y(_y) {}
  void read() { scanf("%I64d%I64d", &x, &y); }
};
Point operator-(Point a, Point b) { return Point(a.x - b.x, a.y - b.y); }
long long Cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
const int N = 222;
const int MOD = (int)1e9 + 7;
int n, dp[N][N];
Point dot[N];
void add(int &a, int b) {
  a += b;
  if (a >= MOD) {
    a -= MOD;
  }
}
int work(int l, int r) {
  if (r - l <= 1) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  int ret = 0;
  for (int i = l + 1; i < r; i++) {
    if (Cross(dot[i] - dot[l], dot[r] - dot[l]) < 0) {
      add(ret, (long long)work(l, i) * work(i, r) % MOD);
    }
  }
  return dp[l][r] = ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    dot[i].read();
  }
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    sum += Cross(dot[i], dot[(i + 1) % n]);
  }
  if (sum > 0) {
    reverse(dot, dot + n);
  }
  memset(dp, -1, sizeof(dp));
  printf("%d\n", work(0, n - 1));
  return 0;
}
