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;
template <class T>
inline void read(T& x) {
  bool fu = 0;
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') fu = 1, c = getchar();
  for (x = 0; c > 32; c = getchar()) x = x * 10 + c - '0';
  if (fu) x = -x;
};
template <class T>
inline void read(T& x, T& y) {
  read(x);
  read(y);
}
template <class T>
inline void read(T& x, T& y, T& z) {
  read(x);
  read(y);
  read(z);
}
inline char getc() {
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  return c;
}
const int N = 210;
const int mod = int(1e9) + 7;
int n, i, j, k, l, p;
int f[N][N];
bool can[N][N][N], no[N][N];
long long s[N][N];
struct node {
  long long x, y;
} a[N];
void read(node& a) { read(a.x, a.y); }
int sgn(long long x) {
  if (x == 0) return 0;
  return x > 0 ? 1 : -1;
}
node operator+(node a, node b) { return (node){a.x + b.x, a.y + b.y}; }
node operator-(node a, node b) { return (node){a.x - b.x, a.y - b.y}; }
node operator*(node a, long long x) { return (node){a.x * x, a.y * x}; }
node operator/(node a, long long x) { return (node){a.x / x, a.y / x}; }
node mul(node a, node b) { return (node){a.x * b.x, a.y * b.y}; }
long long operator*(node a, node b) { return a.x * b.y - a.y * b.x; }
long long xj(node a, node b, node c) { return (b - a) * (c - a); }
bool on(node a, node b, node c) {
  if (a.x >= min(b.x, c.x) && a.x <= max(b.x, c.x) && a.y >= min(b.y, c.y) &&
      a.y <= max(b.y, c.y) && xj(b, a, c) == 0)
    return 1;
  return 0;
}
bool gx(int x, int y, int p) { return xj(a[x], a[y], a[p]) == 0; }
void pre() {
  int i, j, k, x, y, p;
  long long tot, all = 0;
  for (i = 0; i <= n - 1; i++) all += a[i] * a[(i + 1) % n];
  for (i = 0; i <= n - 1; i++) {
    p = i;
    tot = 0;
    for (j = 1; j <= n - 1; j++) {
      tot += a[p] * a[(p + 1) % n];
      p = (p + 1) % n;
      s[i][p] = tot + a[p] * a[i];
    }
  }
  for (i = 0; i <= n - 1; i++)
    for (j = i + 1; j <= n - 1; j++) {
      if (make_pair(i, j) == make_pair(0, 19)) {
        int o = 1;
      }
      for (k = 0; k <= n - 1; k++)
        if (k != i && k != j)
          if (on(a[k], a[i], a[j])) {
            no[i][j] = 1;
          }
    }
  memset(can, 1, sizeof(can));
  for (i = 0; i <= n - 1; i++)
    for (j = i + 1; j <= n - 1; j++)
      for (k = j + 1; k <= n - 1; k++) {
        if (abs(s[i][j]) + abs(s[j][k]) + abs(s[k][i]) +
                abs(a[i] * a[j] + a[j] * a[k] + a[k] * a[i]) !=
            abs(all))
          can[i][j][k] = 0;
        if (no[i][j] || no[i][k] || no[j][k]) can[i][j][k] = 0;
        if (gx(i, j, k)) can[i][j][k] = 0;
      }
}
bool pd(int x, int y, int z) {
  if (x == y || y == z || x == z) {
    printf("no!\n");
    while (1)
      ;
  }
  if (x > y) swap(x, y);
  if (y > z) swap(y, z);
  if (x > y) swap(x, y);
  return can[x][y][z];
}
int dp(int x, int y) {
  if (x + 2 >= y) return 1;
  if (f[x][y] != -1) return f[x][y];
  int z, ans = 0;
  for (z = x + 1; z <= y - 1; z++)
    if (pd(x, y, z)) ans = (ans + 1LL * dp(x, z) * dp(z, y)) % mod;
  return f[x][y] = ans;
}
int main() {
  read(n);
  for (i = 0; i <= n - 1; i++) read(a[i]), a[i] = a[i] * 2;
  pre();
  memset(f, -1, sizeof(f));
  cout << dp(0, n - 1) << endl;
  return 0;
}
