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;
const int maxn = 400 + 5;
const int inf = 1e9;
const int MOD = 1e9 + 7;
const double eps8 = 1e-8;
const double eps9 = 1e-9;
const double pi = acos(-1);
struct pnt {
  double x, y;
  pnt() {}
  pnt(double x, double y) : x(x), y(y) {}
  pnt operator-(pnt A) { return pnt(x - A.x, y - A.y); }
  double operator*(pnt A) { return x * A.y - y * A.x; }
  double operator%(pnt A) { return x * A.x + y * A.y; }
  void read() { scanf("%lf%lf", &x, &y); }
  void Epr() { fprintf(stderr, "x y: %.3lf %.3lf\n", x, y); }
};
pnt p[200];
bool a[200][200];
long long dp[maxn][maxn];
int n;
int next(int x) { return (x + 1) % n; }
bool boundBox(double a, double b, double c, double d) {
  if (a > b) swap(a, b);
  if (c > d) swap(c, d);
  return max(a, c) <= min(b, d);
}
bool intersec(pnt A, pnt B, pnt C, pnt D) {
  return boundBox(A.x, B.x, C.x, D.x) && boundBox(A.y, B.y, C.y, D.y) &&
         ((C - A) * (B - A)) * ((D - A) * (B - A)) <= eps8 &&
         ((A - D) * (C - D)) * ((B - D) * (C - D)) <= eps8;
}
bool check(pnt A, pnt B, pnt C, pnt D) {
  pnt v, u, t;
  v = A - B;
  u = C - B;
  t = D - B;
  if ((v * u) * (v * t) < 0) {
    return v * u > 0;
  }
  if ((v * u) == 0) {
    return t * v > 0;
  }
  if ((v * t) == 0) {
    return v * u > 0;
  }
  return u * t > 0;
}
int main() {
  int v;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) p[i].read();
  double ang = 0;
  for (int i = 0; i < n; i++) {
    pnt v, u;
    v = p[(i + 1) % n] - p[i];
    u = p[(i + 2) % n] - p[(i + 1) % n];
    ang += atan2(v * u, v % u);
  }
  if (ang < 0) {
    reverse(p, p + n);
  }
  for (int i = 0; i < n; i++)
    for (int j = i + 2; j < n; j++) {
      if (i == 0 && j == n - 1) continue;
      bool flag = 1;
      for (int k = 0; k < n; k++) {
        if (i == k || i == (k + 1) % n || j == k || j == (k + 1) % n) continue;
        flag &= !intersec(p[i], p[j], p[k], p[(k + 1) % n]);
      }
      flag &= check(p[(i - 1 + n) % n], p[i], p[(i + 1) % n], p[j]);
      flag &= check(p[(j - 1 + n) % n], p[j], p[(j + 1) % n], p[i]);
      a[i][j] = a[j][i] = flag;
    }
  for (int i = 0; i < n; i++) a[i][(i + 1) % n] = a[(i + 1) % n][i] = 1;
  for (int i = 0; i < n; i++) dp[i][1] = 1;
  for (int i = 0; i < n; i++)
    if (a[i][(i + 2) % n]) dp[i][2] = 1;
  for (int len = 3; len < n; len++) {
    for (int l = 0; l < n; l++) {
      int r = (l + len) % n;
      if (a[l][r] == 0) continue;
      if (a[next(l)][r])
        dp[l][len] = (dp[l][len] + dp[(l + 1) % n][len - 1]) % MOD;
      for (int i = l + 2; i < l + len; i++) {
        v = i % n;
        if (a[l][v] == 0) continue;
        assert(0 <= l && l < 200);
        assert(0 <= len && len < 200);
        assert(0 <= i - l && i - l < 200);
        assert(0 <= l + len - i && l + len - i < 200);
        dp[l][len] = (dp[l][len] + dp[l][i - l] * dp[i][l + len - i]) % MOD;
      }
    }
  }
  cout << dp[0][n - 1] << endl;
  return 0;
}
