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;
inline int read() {
  int x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') positive = 0;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
const int N = 205;
const double eps = 1e-8;
struct node {
  double x, y;
  node() {}
  node(double _x, double _y) : x(_x), y(_y) {}
  node operator+(const node &A) const { return node(x + A.x, y + A.y); }
  node operator-(const node &A) const { return node(x - A.x, y - A.y); }
  node operator*(const double &A) const { return node(x * A, y * A); }
  node operator/(const double &A) const { return node(x / A, y / A); }
  double operator*(const node &A) const { return x * A.x + y * A.y; }
  double operator%(const node &A) const { return x * A.y - y * A.x; }
  double operator/(const node &A) const {
    return fabs(A.x) < eps ? y / A.y : x / A.x;
  }
  bool operator<(const node &A) const {
    return fabs(x - A.x) < eps ? y < A.y : x < A.x;
  }
  double angle() const { return atan2(y, x); }
  double norm() const { return x * x + y * y; }
  double len() const { return sqrt(x * x + y * y); }
  void Read() { x = read(), y = read(); }
  void prt() const { printf("%lf %lf\n", x, y); }
} a[N];
bool b[N][N], v[N][N];
int n, ans, f[N][N];
int chg(double a) { return fabs(a) < eps ? 0 : a > 0 ? 1 : -1; }
bool cross(const node &A, const node &B, const node &C, const node &D) {
  return chg((B - A) % (D - A)) * chg((B - A) % (C - A)) < 0 &&
         chg((D - C) % (A - C)) * chg((D - C) % (B - C)) < 0;
}
bool up(node A, node B, node C) {
  if (A.x > B.x) swap(A, B);
  return A.x < C.x + eps && C.x + eps < B.x && ((B - A) % (C - A) < 0);
}
bool inside(const node &A) {
  bool cnt = 0;
  for (int i = 1; i <= n; ++i)
    if (up(a[i - 1], a[i], A)) cnt ^= 1;
  return cnt;
}
bool check(int l, int r) {
  if (l + 1 == r) return 1;
  for (int i = 1; i <= n; ++i)
    if (i != l && i != r && !chg((a[l] - a[i]) % (a[r] - a[i])) &&
        (a[i] - a[l]) / (a[r] - a[l]) > eps &&
        (a[i] - a[l]) / (a[r] - a[l]) < 1)
      return 0;
  for (int i = 1; i <= n; ++i)
    if (cross(a[l], a[r], a[i - 1], a[i])) return 0;
  return inside((a[l] + a[r]) / 2.);
}
long long dfs(int l, int r) {
  if (l + 1 == r) return 1;
  if (v[l][r]) return f[l][r];
  v[l][r] = 1;
  int &res = f[l][r];
  res = 0;
  for (int k = l + 1; k <= r - 1; ++k)
    if (b[l][k] && b[r][k] && fabs((a[l] - a[k]) % (a[r] - a[k])) > eps)
      res = (res + dfs(l, k) * dfs(k, r)) % 1000000007;
  return res;
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i].Read();
  *a = a[n];
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= i - 1; ++j) b[i][j] = b[j][i] = check(j, i);
  printf("%d\n", (int)dfs(1, n));
  return 0;
}
