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 = 200 + 10;
const int MOD = 1000 * 1000 * 1000 + 7;
struct Point {
  int x, y;
  Point(int x = 0, int y = 0) : x(x), y(y) {}
} p[maxn];
Point operator-(const Point& a, const Point& b) {
  return Point(a.x - b.x, a.y - b.y);
}
long long cross(const Point& a, const Point& b) {
  return (long long)a.x * b.y - (long long)a.y * b.x;
}
int d[maxn][maxn];
int n;
long long solve(int s, int e) {
  int& ans = d[s][e];
  if (ans != -1) return ans;
  if (s + 1 == e) return ans = 1;
  long long res = 0;
  for (int i = s + 1; i < e; ++i) {
    if (cross(p[s] - p[e], p[i] - p[e]) > 0)
      res = (res + solve(s, i) * solve(i, e)) % MOD;
  }
  return ans = res;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d %d", &p[i].x, &p[i].y);
  long long judge = 0;
  for (int i = 0; i < n; ++i) judge += cross(p[i], p[(i + 1) % n]);
  if (judge < 0) reverse(p, p + n);
  memset(d, -1, sizeof(d));
  printf("%lld\n", solve(0, n - 1));
  return 0;
}
