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 INF = 1e9;
const int MAXN = 2e2 + 5;
const int MOD = 1e9 + 7;
int pre[MAXN][MAXN];
long long pd[MAXN][MAXN];
long long cross(pair<long long, long long> a, pair<long long, long long> b,
                pair<long long, long long> c) {
  return ((b.first - a.first) * (c.second - a.second) -
          (b.second - a.second) * (c.first - a.first));
}
pair<long long, long long> v[MAXN];
long long solve(int l, int r) {
  if (r - l == 0) return 0LL;
  if (r - l == 1) return 1LL;
  if (r - l == 2) return (cross(v[l], v[l + 1], v[r]) > 0 ? 1 : 0);
  if (pd[l][r] == -1) {
    long long ans = 0LL;
    for (int x = l + 1; x < r; ++x) {
      if (!pre[l][x]) continue;
      if (!pre[x][r]) continue;
      long long tmp = cross(v[l], v[x], v[r]);
      if (tmp <= 0) continue;
      ans = (ans + solve(l, x) * solve(x, r)) % MOD;
    }
    pd[l][r] = ans;
  }
  return pd[l][r];
}
enum Posicao { ESQUERDA, DIREITA, ORIGEM, DESTINO, ENTRE, ANTES, DEPOIS };
Posicao classify(pair<long long, long long> a, pair<long long, long long> b,
                 pair<long long, long long> c) {
  long long o2d = cross(a, b, c);
  if (o2d > 0)
    return ESQUERDA;
  else if (o2d < 0)
    return DIREITA;
  else if (a.first == c.first && a.second == c.second)
    return ORIGEM;
  else if (b.first == c.first && b.second == c.second)
    return DESTINO;
  else if (a.first != b.first) {
    long long d1 = c.first - a.first, d2 = b.first - c.first;
    if (d1 * d2 > 0)
      return ENTRE;
    else if (abs(d1) < abs(d2))
      return ANTES;
    return DEPOIS;
  }
  long long d1 = c.second - a.second, d2 = b.second - c.second;
  if (d1 * d2 > 0)
    return ENTRE;
  else if (abs(d1) < abs(d2))
    return ANTES;
  return DEPOIS;
}
bool left(pair<long long, long long> a, pair<long long, long long> b,
          pair<long long, long long> c) {
  return cross(a, b, c) > 0;
}
bool intersection(pair<long long, long long> a, pair<long long, long long> b,
                  pair<long long, long long> c, pair<long long, long long> d) {
  int aux = classify(a, b, c);
  if (aux == ORIGEM || aux == ENTRE || aux == DESTINO) return true;
  aux = classify(a, b, d);
  if (aux == ORIGEM || aux == ENTRE || aux == DESTINO) return true;
  aux = classify(c, d, a);
  if (aux == ORIGEM || aux == ENTRE || aux == DESTINO) return true;
  aux = classify(c, d, b);
  if (aux == ORIGEM || aux == ENTRE || aux == DESTINO) return true;
  return ((left(a, b, c) ^ left(a, b, d)) && (left(c, d, a) ^ left(c, d, b)));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  while (cin >> n) {
    for (int i = 0; i < n; ++i) {
      int x, y;
      cin >> x >> y;
      v[i] = pair<long long, long long>(x, y);
    }
    v[n] = v[0];
    memset(pd, -1, sizeof pd);
    long long area = 0LL;
    for (int i = 1; i < n - 1; ++i) area += cross(v[0], v[i], v[i + 1]);
    if (area < 0) {
      reverse(v + 1, v + n);
    }
    for (int i = 0; i < n; ++i) {
      for (int j = i; j < n; ++j) {
        pre[i][j] = 1;
        for (int k = 0; k < n; ++k) {
          if (v[i] == v[k] || v[i] == v[k + 1]) continue;
          if (v[j] == v[k] || v[j] == v[k + 1]) continue;
          if (intersection(v[i], v[j], v[k], v[k + 1])) pre[i][j] = 0;
        }
      }
    }
    cout << solve(0, n - 1) << endl;
  }
}
