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;
int n;
const int mod = 1000000007;
pair<int, int> dat[203];
long long dp[203][203];
long long ccw(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
  b.first -= a.first, b.second -= a.second;
  c.first -= a.first, c.second -= a.second;
  return b.first * (long long)c.second - b.second * (long long)c.first;
}
long long cntTri(int s, int e) {
  if (s + 1 == e) return 1;
  if (dp[s][e] != -1) return dp[s][e];
  long long ans = 0;
  for (int i = s + 1; i < e; i++) {
    if (ccw(dat[s], dat[e], dat[i]) >= 0) continue;
    ans += cntTri(s, i) * cntTri(i, e) % mod;
    ans %= mod;
  }
  return dp[s][e] = ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &dat[i].first, &dat[i].second);
  }
  dat[n] = dat[0];
  long long orientation = 0;
  for (int i = 0; i < n; i++)
    orientation += (dat[i].second + dat[i + 1].second) *
                   (long long)(dat[i].first - dat[i + 1].first);
  if (orientation < 0) reverse(dat, dat + n + 1);
  memset(dp, -1, sizeof(dp));
  long long ans = cntTri(0, n - 1);
  printf("%lld\n", ans);
  return 0;
}
