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 = 2e2 + 7;
const double eps = 1e-8;
const double inf = 1e20;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
int sgn(double x) {
  if (fabs(x) < eps) return 0;
  if (x < 0)
    return -1;
  else
    return 1;
}
inline double square(double x) { return x * x; }
struct Point {
  double x, y;
  Point() = default;
  Point(double x_, double y_) : x(x_), y(y_) {}
  void input() { scanf("%lf%lf", &x, &y); }
  void output() { printf("%.2lf %.2lf\n", x, y); }
  Point operator-(const Point &b) const { return Point(x - b.x, y - b.y); }
  bool operator==(Point b) const {
    return sgn(x - b.x) == 0 && sgn(y - b.y) == 0;
  }
  bool operator<(Point b) const {
    return sgn(x - b.x) == 0 ? sgn(y - b.y) < 0 : x < b.x;
  }
  Point operator+(const Point &b) const { return Point(x + b.x, y + b.y); }
  double operator^(const Point &b) const { return x * b.y - y * b.x; }
  double operator*(const Point &b) const { return x * b.x + y * b.y; }
} p[maxn];
int n;
long long dp[maxn][maxn];
long long dfs(int i, int j) {
  if (dp[i][j] != -1) return dp[i][j];
  if (i + 1 == j) return dp[i][j] = 1;
  long long ans = 0;
  for (int k = i + 1; k < j; k++) {
    if (((p[k] - p[i]) ^ (p[j] - p[i])) > 0) {
      ans = (ans + dfs(i, k) * dfs(k, j) % mod) % mod;
    }
  }
  return dp[i][j] = ans;
}
int main() {
  scanf("%d", &n);
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < n; i++) {
    p[i].input();
  }
  double s = 0;
  for (int i = 0; i < n; i++) {
    s += p[i] ^ p[(i + 1) % n];
  }
  if (s < 0) {
    for (int i = 0, j = n - 1; i < j; i++, j--) {
      swap(p[i], p[j]);
    }
  }
  printf("%lld\n", dfs(0, n - 1));
  return 0;
}
