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(long long x = 0, long long y = 0) : x(x), y(y) {}
  point operator+(point &p) { return point(x + p.x, y + p.y); }
  point operator-(point &p) { return point(x - p.x, y - p.y); }
};
int sign(long long x) {
  if (x > 0) return 1;
  if (x < 0) return -1;
  return 0;
}
long long cross(point u, point v) { return u.x * v.y - u.y * v.x; }
long long cross(point a, point b, point c) { return cross(a - b, c - b); }
bool intersect1(point a, point b, point c, point d) {
  return sign(cross(a, b, c)) != sign(cross(a, b, d));
}
bool intersect(point a, point b, point c, point d) {
  return intersect1(a, b, c, d) && intersect1(c, d, a, b);
}
int n;
point p[205];
long long dp[205][205];
long long cross(int a, int b, int c) { return cross(p[a], p[b], p[c]); }
void make_ccw() {
  long long a = 0;
  for (int i = 1; i <= n; i++) a += cross(p[i], p[i + 1]);
  if (a < 0) reverse(p, p + n + 2);
}
bool good(int a, int b) {
  if (a == 1 && b == n) return 1;
  if (cross(a - 1, a, a + 1) <= 0 && cross(a - 1, a, b) >= 0) return 0;
  if (cross(a - 1, a, a + 1) <= 0 && cross(a + 1, a, b) <= 0) return 0;
  if (cross(a - 1, a, a + 1) >= 0 && cross(a - 1, a, b) >= 0 &&
      cross(a + 1, a, b) <= 0)
    return 0;
  for (int i = 1; i <= n; i++) {
    if (a == i || b == i || a == i + 1 || b == i + 1 || a == i - n + 1 ||
        b == i - n + 1)
      continue;
    if (intersect(p[a], p[b], p[i], p[i + 1])) return 0;
  }
  return 1;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> p[i].x >> p[i].y;
  p[0] = p[n];
  p[n + 1] = p[1];
  make_ccw();
  for (int i = 1; i < n; i++) dp[i][i + 1] = 1;
  for (int i = 2; i < n; i++) {
    for (int j = 1; j + i <= n; j++) {
      if (!good(j, j + i)) continue;
      for (int k = j + 1; k < j + i; k++) {
        dp[j][j + i] += dp[j][k] * dp[k][j + i] % 1000000007;
        dp[j][j + i] %= 1000000007;
      }
    }
  }
  cout << dp[1][n] << '\n';
}
