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 n;
const int MOD = 7 + (int)1e9;
struct point {
  long long x, y;
  void read() {
    int xx, yy;
    scanf("%d%d", &xx, &yy);
    x = xx;
    y = yy;
  }
  point operator-(point a) { return (point){x - a.x, y - a.y}; }
} P[300];
long long det(point x, point y) { return x.x * y.y - x.y * y.x; }
int can[300][300];
bool sig(long long x) {
  if (x > 0) return 1;
  if (x < 0) return -1;
  return 0;
}
bool chk(point a, point b, point c, point d) {
  point A = b - a, B = c - a, C = d - a;
  long long sig1 = det(A, B), sig2 = det(A, C);
  sig1 = sig(sig1);
  sig2 = sig(sig2);
  if (sig1 * sig2 != -1) return 0;
  A = d - c, B = a - c, C = b - c;
  sig1 = det(A, B), sig2 = det(A, C);
  sig1 = sig(sig1);
  sig2 = sig(sig2);
  if (sig1 * sig2 != -1) return 0;
  return 1;
}
bool check(int x, int y) {
  for (int i = 1; i < n; i++) {
    if (chk(P[x], P[y], P[i], P[i + 1])) return 0;
  }
  if (chk(P[x], P[y], P[1], P[n])) return 0;
  return 1;
}
int f[300][300];
int solve(int l, int r) {
  if (l > r) return 0;
  if (f[l][r] != -1) return f[l][r];
  if (l + 1 >= r) return 1;
  if (l + 2 == r) {
    if (det(P[l], P[l + 1]) + det(P[l + 1], P[r]) + det(P[r], P[l]) > 0) {
      if (can[l][l + 1] && can[l + 1][r] && can[l][r])
        return 1;
      else
        return 0;
    } else
      return 0;
  }
  int res = 0;
  for (int sel = l + 1; sel < r; sel++) {
    if (det(P[l], P[sel]) + det(P[sel], P[r]) + det(P[r], P[l]) > 0) {
      if (can[l][sel] && can[sel][r] && can[l][r]) {
        int x = solve(l, sel), y = solve(sel, r);
        x = (long long)x * y % MOD;
        (res += x) %= MOD;
      }
    }
  }
  return f[l][r] = res;
}
int main() {
  memset(f, -1, sizeof f);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) P[i].read();
  long long area = det(P[n], P[1]);
  for (int i = 1; i < n; i++) area += det(P[i], P[i + 1]);
  if (area < 0) reverse(P + 1, P + 1 + n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i != j) can[i][j] = check(i, j);
  printf("%d\n", solve(1, n));
}
