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 MAXN = 210;
const long long MOD = 1000000007;
struct Point {
  long long x, y;
  Point() {}
  Point(long long x, long long y) : x(x), y(y) {}
  Point operator+(const Point& p) const { return Point(x + p.x, y + p.y); }
  Point operator-(const Point& p) const { return Point(x - p.x, y - p.y); }
};
Point vertex[MAXN];
long long dp[MAXN][MAXN];
long long Cross(const Point& A, const Point& B) {
  return A.x * B.y - A.y * B.x;
}
bool IsClockwise(int n) {
  long long ret = 0;
  for (int i = 0; i < n; ++i) {
    ret += Cross(vertex[i], vertex[(i + 1) % n]);
  }
  return ret < 0;
}
long long Gao(int l, int r) {
  if (dp[l][r] != -1) {
    return dp[l][r];
  }
  long long& ret = dp[l][r] = 0;
  if (l + 1 == r) {
    return ret = 1;
  }
  for (int k = l + 1; k < r; ++k) {
    if (Cross(vertex[r] - vertex[l], vertex[k] - vertex[l]) > 0) {
      ret = (ret + Gao(l, k) * Gao(k, r)) % MOD;
    }
  }
  return ret;
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < n; ++i) {
      long long x, y;
      scanf("%I64d%I64d", &x, &y);
      vertex[i] = Point(x, y);
    }
    if (!IsClockwise(n)) {
      reverse(vertex, vertex + n);
    }
    memset(dp, -1, sizeof(dp));
    printf("%I64d\n", Gao(0, n - 1));
  }
  return 0;
}
