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>
const int N = 209, MOD = 1000000007;
long long f[N][N], area[N][N];
struct point {
  long long x, y;
  point(const long long _x = 0, const long long _y = 0) : x(_x), y(_y) {}
} p[N];
int n;
inline int sign(long long x) { return x < 0 ? -1 : x > 0; }
inline long long cpr(const point &a, const point &b, const point &c) {
  return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
}
inline bool cross(point &u, point &v, point &a, point &b) {
  if (std::max(u.x, v.x) < std::min(a.x, b.x) ||
      std::min(u.x, v.x) > std::max(a.x, b.x))
    return false;
  if (std::max(u.y, v.y) < std::min(a.x, b.y) ||
      std::min(u.y, v.y) > std::max(a.y, b.y))
    return false;
  if (sign(cpr(u, v, a)) * sign(cpr(u, v, b)) <= 0 &&
      sign(cpr(a, b, u)) * sign(cpr(a, b, v)) <= 0)
    return true;
  return false;
}
bool check(int i, int j) {
  if (i + 1 == j || i == 1 && j == n) return true;
  if (sign(cpr(p[i], p[j], p[i + 1])) * sign(cpr(p[i], p[j], p[i - 1])) > 0)
    return false;
  for (int k = 1; k <= n; ++k)
    if (k != i && k % n + 1 != i && k != j && k % n + 1 != j) {
      if (cross(p[i], p[j], p[k], p[k + 1])) return false;
    } else if (!cpr(p[i], p[j], p[(k == i || k == j) ? k + 1 : k]))
      return false;
  return true;
}
long long F(int i, int j) {
  if (f[i][j] != -1) return f[i][j];
  if (i + 1 == j) return 1;
  f[i][j] = 0;
  for (int k = i + 1; k < j; ++k)
    if (cpr(p[i], p[j], p[k]) > 0) f[i][j] += F(i, k) * F(k, j) % MOD;
  f[i][j] %= MOD;
  return f[i][j];
}
int main() {
  memset(f, -1, sizeof f);
  scanf("%d", &n);
  long long S = 0;
  for (int i = 1; i <= n; ++i) scanf("%I64d%I64d", &p[i].x, &p[i].y);
  p[n + 1] = p[1];
  p[0] = p[n];
  for (int i = 1; i <= n; ++i) S += cpr(point(0, 0), p[i], p[i + 1]);
  if (S > 0) std::reverse(p + 1, p + 1 + n);
  p[n + 1] = p[1];
  p[0] = p[n];
  printf("%I64d\n", F(1, n));
}
