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;
pair<int, int> operator+(const pair<int, int> a, const pair<int, int> b) {
  return {a.first + b.first, a.second + b.second};
}
pair<int, int> operator-(const pair<int, int> a, const pair<int, int> b) {
  return {a.first - b.first, a.second - b.second};
}
pair<int, int> operator*(const pair<int, int> a, const int b) {
  return {a.first * b, a.second * b};
}
pair<int, int> operator/(const pair<int, int> a, const int b) {
  return {a.first / b, a.second / b};
}
pair<int, int> operator&(const pair<int, int> a, const pair<int, int> b) {
  return {a.first * b.first - a.second * b.second,
          a.first * b.second + a.second * b.first};
}
long long operator^(const pair<int, int> a, const pair<int, int> b) {
  return (long long)a.first * b.second - (long long)a.second * b.first;
}
long long operator*(const pair<int, int> a, const pair<int, int> b) {
  return (long long)a.first * b.first + (long long)a.second * b.second;
}
bool operator!=(const pair<int, int> a, const pair<int, int> b) {
  return a.first == b.first ? a.second != b.second : 1;
}
pair<int, int> conj(pair<int, int> a) { return {-a.second, a.first}; }
void Input(pair<int, int> &a) { scanf("%d%d", &a.first, &a.second); }
void Print(pair<int, int> a) { cout << a.first << ' ' << a.second << endl; }
int sgn(long long a) { return (a > 0ll) - (a < 0ll); }
const int mod = 1e9 + 7;
const int NN = 222;
pair<int, int> p[NN];
bool is[NN][NN];
int dp[NN][NN];
bool is_on(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
  if (sgn(a - b ^ b - c)) return 0;
  return sgn((a - c) * (b - c)) <= 0;
}
bool is_in_poly(pair<int, int> a, pair<int, int> *p, int n) {
  for (int i = 0; i < n; i++)
    if (is_on(p[i], p[i + 1], a)) return 1;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    pair<int, int> b = p[i], c = p[i + 1];
    if (b > c) swap(b, c);
    if (sgn(c - b ^ a - b) < 0 && b.first < a.first && a.first <= c.first)
      cnt++;
  }
  return cnt & 1;
}
int main() {
  int n, i, j, k, l, ii, jj, kk, ll;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    Input(p[i]);
    p[i] = p[i] * 2;
  }
  p[n] = p[0];
  long long area = 0;
  for (i = 0; i < n; i++) area += p[i] ^ p[i + 1];
  if (area < 0) reverse(p, p + n + 1);
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      if (i == j) continue;
      is[i][j] = true;
      if (abs(i - j) == 1 || abs(i - j) == n - 1) {
        continue;
      }
      pair<int, int> q = (p[i] + p[j]) / 2;
      if (!is_in_poly(q, p, n)) {
        is[i][j] = false;
        continue;
      }
      for (k = 0; k < n; k++) {
        if (k != i && k != j && is_on(p[i], p[j], p[k])) {
          is[i][j] = false;
          break;
        }
        ii = sgn(p[k] - p[i] ^ p[j] - p[i]);
        jj = sgn(p[k + 1] - p[i] ^ p[j] - p[i]);
        kk = sgn(p[i] - p[k] ^ p[k + 1] - p[k]);
        ll = sgn(p[j] - p[k] ^ p[k + 1] - p[k]);
        if (ii && jj && kk && ll && ii != jj && kk != ll) {
          is[i][j] = false;
          break;
        }
      }
    }
  }
  for (i = 0; i < n; i++) dp[i][(i + 1) % n] = 1;
  for (k = 2; k < n; k++) {
    for (i = 0; i < n; i++) {
      jj = (i + k) % n;
      if (!is[i][jj]) continue;
      for (j = i + 1; j < i + k; j++) {
        if (is[i][j % n] && is[j % n][jj] &&
            sgn(p[j % n] - p[i] ^ p[jj] - p[i])) {
          dp[i][jj] += (long long)dp[i][j % n] * dp[j % n][jj] % mod;
          if (dp[i][jj] >= mod) dp[i][jj] -= mod;
        }
      }
    }
  }
  cout << dp[0][n - 1] << endl;
  return 0;
}
