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 MOD = 1000000007;
struct Vec {
  long long x, y;
  Vec(long long _x, long long _y) : x(_x), y(_y) {}
  long long operator*(const Vec &that) const { return x * that.y - y * that.x; }
  Vec operator-(const Vec &that) const { return Vec(x - that.x, y - that.y); }
};
vector<Vec> pt;
long long area(vector<Vec> &pt) {
  long long res = 0;
  for (int i = 0; i < pt.size(); i++) {
    res += pt[i] * pt[(i + 1) % pt.size()];
  }
  return res;
}
bool clock(Vec a, Vec b, Vec c) {
  b = b - a;
  c = c - a;
  return b * c < 0;
}
int n;
long long dp[211][211];
long long dpr(int l, int r) {
  long long &res = dp[l][r];
  if (res != -1)
    return res;
  else if (r - l == 1)
    return res = 1;
  res = 0;
  for (int i = l + 1; i < r; i++) {
    if (clock(pt[l], pt[r], pt[i])) {
      res += dpr(l, i) * dpr(i, r);
      res %= MOD;
    }
  }
  return res;
}
void input() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    pt.push_back(Vec(x, y));
  }
}
void solve() {
  if (area(pt) < 0) reverse(pt.begin(), pt.end());
  memset(dp, -1, sizeof(dp));
  printf("%I64d\n", dpr(0, n - 1));
}
int main() {
  input();
  solve();
}
