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;
int mod = 1000000007;
int n, x[202], y[202];
bool f[202][202];
int sgn(long long x) { return (x > 0) - (x < 0); }
int cw(int a, int b, int c) {
  return sgn(1LL * (x[b] - x[a]) * (y[c] - y[a]) -
             1LL * (y[b] - y[a]) * (x[c] - x[a]));
}
bool has(int a, int b, int c, int d) {
  return cw(a, b, c) * cw(a, b, d) < 0 && cw(c, d, a) * cw(c, d, b) < 0;
}
double dist(int a, int b) {
  return sqrt(1.0 * (x[a] - x[b]) * (x[a] - x[b]) +
              1.0 * (y[a] - y[b]) * (y[a] - y[b]));
}
bool on(int a, int b, int c) {
  return cw(a, b, c) == 0 && fabs(dist(a, b) - dist(a, c) - dist(b, c)) < 1e-7;
}
long long calcS(const vector<pair<int, int> >& v) {
  long long s = 0;
  for (int i = 0; i < v.size(); ++i) {
    s += 1LL * v[i].first * v[(i + 1) % v.size()].second -
         1LL * v[i].second * v[(i + 1) % v.size()].first;
  }
  return s > 0 ? s : -s;
}
bool isInside(int a, int b) {
  vector<pair<int, int> > v;
  for (int i = 0; i < n; ++i) v.push_back(make_pair(x[i], y[i]));
  long long s = calcS(v);
  v.clear();
  for (int i = a;; ++i) {
    if (i == n) i = 0;
    v.push_back(make_pair(x[i], y[i]));
    if (i == b) break;
  }
  s -= calcS(v);
  v.clear();
  for (int i = b;; ++i) {
    if (i == n) i = 0;
    v.push_back(make_pair(x[i], y[i]));
    if (i == a) break;
  }
  s -= calcS(v);
  return s == 0;
}
bool ok(int a, int b) {
  bool ret = true;
  if ((a + 1) % n != b && (b + 1) % n != a) {
    for (int i = 0; i < n; ++i) {
      if (a != i && a != (i + 1) % n && b != i && b != (i + 1) % n) {
        if (has(a, b, i, i + 1)) {
          ret = false;
          break;
        }
      }
    }
    if (!isInside(a, b)) ret = false;
  }
  for (int i = 0; i < n; ++i) {
    if (i != a && i != b && on(a, b, i)) ret = false;
  }
  return f[a][b] = f[b][a] = ret;
}
int dp[202][202];
int nx(int i) {
  if (i + 1 < n) return i + 1;
  return 0;
}
int solve(int l, int r) {
  if (dp[l][r] != -1) return dp[l][r];
  if (l == r) return 1;
  if (nx(l) == r) return dp[l][r] = 1;
  int ret = 0;
  for (int i = nx(l);; i = nx(i)) {
    if (i == r) break;
    if (f[l][i] && f[r][i]) {
      ret = (ret + 1LL * solve(l, i) * solve(i, r)) % mod;
    }
  }
  return dp[l][r] = ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &x[i], &y[i]);
  }
  x[n] = x[0];
  y[n] = y[0];
  for (int i = 0; i < n; ++i) {
    f[i][i] = 0;
    for (int j = i + 1; j < n; ++j) {
      ok(i, j);
    }
  }
  int ans = 0;
  memset(dp, -1, sizeof(dp));
  ans = solve(0, n - 1);
  printf("%d\n", ans);
  return 0;
}
