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 N = 200 + 5;
template <typename T>
bool cmax(T &a, T b) {
  return (a < b) ? a = b, 1 : 0;
}
template <typename T>
bool cmin(T &a, T b) {
  return (a > b) ? a = b, 1 : 0;
}
template <typename T>
T read() {
  T ans = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') f = -1, ch = getchar();
  while (isdigit(ch))
    ans = (ans << 3) + (ans << 1) + (ch - '0'), ch = getchar();
  return ans * f;
}
template <typename T>
void write(T x, char y) {
  if (x == 0) {
    putchar('0'), putchar(y);
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  static char wr[20];
  int top = 0;
  for (; x; x /= 10) wr[++top] = x % 10 + '0';
  while (top) putchar(wr[top--]);
  putchar(y);
}
void file() {}
struct node {
  int x, y;
  node operator-(const node b) const { return (node){x - b.x, y - b.y}; }
  long long operator*(const node b) const {
    return 1ll * x * b.y - 1ll * b.x * y;
  }
} e[N];
int n;
void input() {
  n = read<int>();
  for (register int i = (int)(0); i <= (int)(n - 1); ++i)
    e[i].x = read<int>(), e[i].y = read<int>();
}
const int mo = 1e9 + 7;
int dp[N][N];
int DP(int l, int r) {
  if (dp[l][r] != -1) return dp[l][r];
  if (l + 1 == r) return dp[l][r] = 1;
  int res = 0;
  for (register int mid = (int)(l + 1); mid <= (int)(r - 1); ++mid) {
    if ((e[mid] - e[l]) * (e[r] - e[l]) > 0)
      res = (res + 1ll * DP(l, mid) * DP(mid, r) % mo) % mo;
  }
  return dp[l][r] = res;
}
void work() {
  memset(dp, -1, sizeof dp);
  long long s = 0;
  for (register int i = (int)(0); i <= (int)(n - 1); ++i)
    s += e[i] * e[(i + 1) % n];
  if (s < 0) reverse(e, e + n);
  write(DP(0, n - 1), '\n');
}
int main() {
  file();
  input();
  work();
  return 0;
}
