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() : x(0), y(0) {}
  point(long long x, long long y) : 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); }
  point operator*(long long k) { return point(x * k, y * k); }
  point operator/(long long k) { return point(x / k, y / k); }
};
inline int sgn(long long a) { return a < 0 ? -1 : a > 0 ? 1 : 0; }
inline long long cross(point a, point b, point c) {
  point u = a - b, v = c - b;
  return u.x * v.y - u.y * v.x;
}
inline int ccw(point a, point b, point c) { return sgn(cross(c, b, a)); }
inline int contains(point a, point b, point c) {
  if (ccw(a, b, c) != 0) return 0;
  return min(a.x, b.x) <= c.x && c.x <= max(a.x, b.x) && min(a.y, b.y) <= c.y &&
         c.y <= max(a.y, b.y);
}
inline int intersect(point a, point b, point c, point d) {
  int s1 = ccw(a, b, c), s2 = ccw(a, b, d);
  int t1 = ccw(c, d, a), t2 = ccw(c, d, b);
  if (s1 != s2 && t1 != t2) return 1;
  if (contains(a, b, c) || contains(a, b, d)) return 1;
  if (contains(c, d, a) || contains(c, d, b)) return 1;
  return 0;
}
inline int in(point a, vector<point> p) {
  int n = p.size(), res = 0;
  for (int i = 0; i < n; i++) {
    point b = p[i], c = p[(i + 1) % n];
    if (contains(b, c, a)) return 1;
    if (min(b.x, c.x) <= a.x && a.x < max(b.x, c.x)) {
      res ^= ((c.x - b.x) * ccw(a, b, c) >= 0);
    }
  }
  return res;
}
const long long MOD = 1000000007;
const int MAXN = 205;
int n, ok[MAXN][MAXN], vis[MAXN][MAXN] = {};
long long dp[MAXN][MAXN] = {};
vector<point> p;
long long check(int a, int b) {
  if (vis[a][b]) return dp[a][b];
  int d = (b - a + n) % n;
  if (d == 1) return 1;
  vis[a][b] = 1;
  for (int i = 1; i < d; i++) {
    int c = (a + i) % n;
    if (!ok[a][c] || !ok[b][c]) continue;
    dp[a][b] = (dp[a][b] + check(a, c) * check(c, b) % MOD) % MOD;
  }
  return dp[a][b];
}
void init() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    point a;
    cin >> a.x >> a.y;
    p.push_back(a + a);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      ok[i][j] = 1;
      if (i == j) ok[i][j] = 0;
      for (int k = 0; k < n; k++) {
        if (i == k || j == k) continue;
        ok[i][j] &= !contains(p[i], p[j], p[k]);
        int l = (k + 1) % n;
        if (i == l || j == l) continue;
        ok[i][j] &= !intersect(p[i], p[j], p[k], p[l]);
      }
      ok[i][j] &= in((p[i] + p[j]) / 2, p);
    }
  }
}
int main() {
  init();
  cout << check(1, 0) << '\n';
}
