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 C>
void mini(C& a4, C b4) {
  a4 = min(a4, b4);
}
template <class C>
void maxi(C& a4, C b4) {
  a4 = max(a4, b4);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
  return out << "(" << pair.first << ", " << pair.second << ")";
}
vector<complex<long long> > poly;
int n;
long long area[300][300];
long long go_area(int start, int stop) {
  long long res = 0;
  int i = start;
  do {
    res += imag(poly[i] * conj(poly[i + 1]));
    i = (i + 1) % n;
  } while (i != stop);
  res += imag(poly[stop] * conj(poly[start]));
  return res;
};
long long dp[300][300];
long long cnt(int a, int b);
long long go_and_try(int a, int x, int b) {
  complex<long long> p1 = poly[x] - poly[a];
  complex<long long> p2 = poly[b] - poly[a];
  long long ta = imag(p1 * conj(p2));
  if (abs(ta) == 0) return 0;
  if (abs(ta) + abs(area[a][x]) + abs(area[x][b]) != abs(area[a][b])) {
    return 0;
  };
  return (cnt(a, x) * cnt(x, b)) % 1000000007;
};
long long cnt(int a, int b) {
  if (dp[a][b] != -1) return dp[a][b];
  if (b == (a + 1) % n) return dp[a][b] = 1;
  long long res = 0;
  for (int i = a + 1; i != b; i = (i + 1) % n) {
    long long cc = go_and_try(a, i, b);
    res = (res + cc) % 1000000007;
  };
  return dp[a][b] = res;
};
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    poly.push_back(complex<long long>(a, b));
  };
  poly.push_back(poly[0]);
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (n); ++j) dp[i][j] = -1;
  long long ar = go_area(0, 0);
  if (ar < 0) {
    reverse((poly).begin(), (poly).end());
  };
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (n); ++j) area[i][j] = go_area(i, j);
  long long rres = cnt(0, n - 1);
  cout << rres << endl;
  return 0;
}
