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;
int getint() {
  int s = 0;
  char c;
  int o = 1;
  for (c = getchar(); c < '0' || c > '9'; c = getchar())
    if (c == '-') o = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) s *= 10, s += c - '0';
  return s * o;
}
const int maxn = 500, mod = (int)1e9 + 7;
struct node {
  int x, y;
} a[maxn];
int n;
long long f[maxn][maxn], sum;
long long b[maxn][maxn];
long long area(node a, node b, node c) {
  long long x1 = b.x - a.x, x2 = c.x - a.x, y1 = b.y - a.y, y2 = c.y - a.y;
  return x1 * y2 - x2 * y1;
}
int check(int s, int t) {
  node b = a[s], c = a[t];
  for (int i = 0; i < n; i++) {
    if (i == s || (i + 1) % n == s || i == t || (i + 1) % n == t) continue;
    if (area(a[i], a[i + 1], b) * area(a[i], a[i + 1], c) <= 0 &&
        area(b, c, a[i]) * area(b, c, a[i + 1]) <= 0)
      return 0;
  }
  long long s1 = 0, s2 = 0;
  for (int i = s; i < t; i++) s1 += area(a[s], a[i], a[i + 1]);
  s1 = abs(s1);
  for (int i = t; i < n + s; i++) s2 += area(a[t], a[i], a[i + 1]);
  s2 = abs(s2);
  return (s1 + s2 == sum);
}
long long dp(int l, int r) {
  if (f[l][r] > -1) return f[l][r];
  if (l + 1 == r) return f[l][r] = 1;
  long long s;
  f[l][r] = 0;
  for (int i = l + 1; i <= r - 1; i++)
    if ((s = area(a[l], a[r], a[i])) && abs(s) + b[l][i] + b[i][r] == b[l][r]) {
      f[l][r] += dp(l, i) * dp(i, r) % mod;
    }
  return f[l][r] %= mod;
}
int main(int argc, char const *argv[]) {
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d%d", &a[i].x, &a[i].y);
  for (int i = 0; i < n; i++) a[i + n] = a[i];
  for (int i = 0; i < n; i++)
    for (int j = i + 2; j < n; j++) {
      b[i][j] = 0;
      for (int k = i + 1; k < j; k++) b[i][j] += area(a[i], a[k], a[k + 1]);
      b[i][j] = abs(b[i][j]);
    }
  memset(f, -1, sizeof(f));
  printf("%d\n", dp(0, n - 1));
  return 0;
}
