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 inf = 1e9;
const double eps = 1e-9;
const double INF = inf;
const double EPS = eps;
const int mod = inf + 7;
inline void madd(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int mul(int a, int b) { return (a * 1ll * b) % mod; }
pair<int, int> operator-(const pair<int, int>& a, const pair<int, int>& b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
long long operator^(const pair<int, int>& a, const pair<int, int>& b) {
  return a.first * 1ll * b.second - a.second * 1ll * b.first;
}
int n;
int D[210][210];
pair<int, int> P[410];
long long S[210][210][210];
inline bool check(int a, int b, int c) {
  int i;
  long long s = S[a][b][c];
  for (i = 0; i < a; i++)
    if (S[a][b][i] + S[b][c][i] + S[c][a][i] == s) return 0;
  for (i = a + 1; i < b; i++)
    if (S[a][b][i] + S[b][c][i] + S[c][a][i] == s) return 0;
  for (i = b + 1; i < c; i++)
    if (S[a][b][i] + S[b][c][i] + S[c][a][i] == s) return 0;
  for (i = c + 1; i < n; i++)
    if (S[a][b][i] + S[b][c][i] + S[c][a][i] == s) return 0;
  return 1;
}
int main() {
  int i, j, len, k;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d%d", &P[i].first, &P[i].second);
  for (i = n - 1; i >= 0; i--) P[i] = P[i] - P[0];
  long long s = 0;
  for (i = 1; i < n - 1; i++) s += (P[i] ^ P[i + 1]);
  if (s < 0) reverse(P + 1, P + n);
  for (i = 0; i < n; i++) P[i + n] = P[i];
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++)
      for (k = 0; k < n; k++) S[i][j][k] = abs((P[i] - P[k]) ^ (P[j] - P[k]));
  for (i = 0; i < n - 1; i++) D[i][i + 1] = 1;
  for (len = 2; len < n; len++)
    for (i = 0; i + len < n; i++) {
      for (j = i + 1; j < i + len; j++) {
        if (check(i, j, i + len))
          madd(D[i][i + len], mul(D[i][j], D[j][i + len]));
      }
    }
  printf("%d\n", D[0][n - 1]);
  ;
  return 0;
}
