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 = 205;
const int MOD = 1000000007;
int n, dp[N][N];
pair<long long, long long> p[N];
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);
}
long long dist(pair<long long, long long> a, pair<long long, long long> b,
               pair<long long, long long> c) {
  return (b.first - a.first) * (c.first - a.first) +
         (b.second - a.second) * (c.second - a.second);
}
long long s(long long x) { return x < 0 ? -1 : (x > 0 ? 1 : 0); }
bool does_intersect(pair<long long, long long> a, pair<long long, long long> b,
                    pair<long long, long long> c, pair<long long, long long> d,
                    bool& flag) {
  long long v1 = s(cross(a, b, c)) * s(cross(a, b, d)),
            v2 = s(cross(c, d, a)) * s(cross(c, d, b));
  flag |= (v1 == 0 && v2 <= 0) || (v1 <= 0 && v2 == 0);
  return v1 < 0 && v2 < 0;
}
bool point_in_polygon(pair<long long, long long> a) {
  int cnt = 0, z = 0;
  bool flag = false;
  do {
    z++;
    flag = false;
    cnt = 0;
    for (int i = 0; i < n; i++) {
      if (does_intersect(a, make_pair(4e8, 4e8 + z), p[i], p[(i + 1) % n],
                         flag)) {
        cnt++;
      }
    }
  } while (flag);
  return cnt % 2 == 1;
}
bool lies_on_segment(pair<long long, long long> a, pair<long long, long long> b,
                     pair<long long, long long> c) {
  return cross(a, b, c) == 0 && dist(a, b, c) > 0 &&
         dist(a, b, c) < dist(a, b, b);
}
bool tmp;
int main() {
  while (cin >> n) {
    for (int i = 0; i < n; i++) {
      cin >> p[i].first >> p[i].second;
      p[i].first *= 2;
      p[i].second *= 2;
    }
    for (int gap = 0; gap < n; gap++)
      if (gap)
        for (int i = 0; i < n; i++) {
          int j = i + gap;
          if (j >= n) break;
          dp[i][j] = 0;
          bool ok = true;
          if (i + 1 == j) {
            dp[i][j] = 1;
          } else {
            for (int x = 0; x < n; x++)
              if (does_intersect(p[i], p[j], p[x], p[(x + 1) % n], tmp))
                ok = false;
            for (int x = 0; x < n; x++)
              if (x != i && x != j && lies_on_segment(p[i], p[j], p[x]))
                ok = false;
            if (ok) {
              if ((i != 0 || j != n - 1) &&
                  !point_in_polygon(make_pair((p[i].first + p[j].first) / 2,
                                              (p[i].second + p[j].second) / 2)))
                ok = false;
              if (ok) {
                for (int k = i + 1; k <= j - 1; k++) {
                  dp[i][j] =
                      (dp[i][j] + ((long long)dp[i][k]) * dp[k][j]) % MOD;
                }
              }
            }
          }
        }
    cout << dp[0][n - 1] << endl;
  }
}
