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;
template <class T>
inline T bigmod(T p, T e, T M) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    T t = bigmod(p, e / 2, M);
    return (t * t) % M;
  }
  return (bigmod(p, e - 1, M) * p) % M;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
struct point {
  long long x, y;
  point() {}
  point(long long _, long long __) {
    x = _;
    y = __;
  }
};
point arr[1000009];
bool clockwise(int n) {
  long long b = 0ll;
  for (int i = 0; i < n; i++) {
    b += (arr[(i + 1) % n].x - arr[i].x) * (arr[(i + 1) % n].y + arr[i].y);
  }
  return b > 0ll;
}
long long product(point a, point b) { return (b.x - a.x) * (b.y + a.y); }
bool isleft(point a, point b, point c) {
  long long sum = product(a, b) + product(b, c) + product(c, a);
  if (sum < 0) return true;
  return false;
}
long long dp[202][202];
long long solve(int l, int r) {
  if (l + 1 == r) return 1ll;
  if (l == r) return 0;
  long long &ret = dp[l][r];
  if (ret != -1ll) return ret;
  ret = 0ll;
  for (int i = l + 1; i < r; i++) {
    if (isleft(arr[l], arr[i], arr[r])) {
      ret = (ret + solve(l, i) * solve(i, r)) % 1000000007;
    }
  }
  return ret;
}
int main() {
  int i, j, k;
  string str, inp;
  int t, n;
  int cases = 1;
  int num, m;
  int a, b, c;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    cin >> arr[i].x >> arr[i].y;
  }
  if (clockwise(n)) reverse(arr, arr + n);
  memset(dp, -1ll, sizeof dp);
  long long res = solve(0, n - 1);
  cout << res << endl;
  return 0;
}
