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 maxn = 200 + 10;
const int mod = 1000000007;
struct Point {
  int x, y;
  Point() {}
  Point(int x, int y) : x(x), y(y) {}
} pt[maxn];
Point dec(Point a, Point b) { return Point(a.x - b.x, a.y - b.y); }
long long Cross(Point a, Point b) {
  return (long long)a.x * b.y - (long long)a.y * b.x;
}
long long dp[maxn][maxn];
long long f(int l, int r) {
  if (r - l <= 1) return 1;
  if (dp[l][r] != -1) return dp[l][r];
  dp[l][r] = 0;
  for (int i = l + 1; i < r; ++i) {
    if (Cross(dec(pt[r], pt[l]), dec(pt[l], pt[i])) < 0)
      dp[l][r] = (dp[l][r] + f(l, i) * f(i, r)) % mod;
  }
  return dp[l][r];
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> pt[i].x >> pt[i].y;
  long long dir = 0;
  pt[n] = pt[0];
  for (int i = 0; i < n; ++i) dir += Cross(pt[i], pt[i + 1]);
  if (dir > 0) reverse(pt, pt + n);
  memset(dp, 0xff, sizeof(dp));
  cout << f(0, n - 1) << endl;
  return 0;
}
