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;
struct Point {
  long long x, y;
  Point() {}
  Point(long long x, long long y) : x(x), y(y) {}
  friend Point operator-(const Point &a, const Point &b) {
    return Point(a.x - b.x, a.y - b.y);
  }
} p[210];
long long Cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
int n;
void read() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> p[i].x >> p[i].y;
}
int Dp() {
  static int dp[210][210];
  static long long area[210][210];
  static bool valid[210][210][210];
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j) {
      for (int k = i + 1; k < j; ++k)
        area[i][j] += Cross(p[k] - p[i], p[k + 1] - p[i]);
      area[i][j] = abs(area[i][j]);
    }
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j) area[j][i] = area[1][n] - area[i][j];
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j)
      for (int k = j + 1; k <= n; ++k)
        if (abs(Cross(p[k] - p[i], p[j] - p[i])) + area[k][i] + area[i][j] +
                    area[j][k] ==
                area[1][n] &&
            Cross(p[k] - p[i], p[j] - p[i]))
          valid[i][j][k] = true;
  for (int i = 1; i < n; ++i) dp[i][i + 1] = 1;
  for (int i = n; i >= 1; --i)
    for (int j = i + 2; j <= n; ++j)
      for (int k = i + 1; k <= j - 1; ++k)
        if (valid[i][k][j])
          dp[i][j] = (dp[i][j] + (long long)dp[i][k] * dp[k][j]) % 1000000007;
  return dp[1][n];
}
int main() {
  read();
  printf("%d\n", Dp());
  return 0;
}
