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 <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T& val) {
  std::fill((T*)array, (T*)(array + N), val);
}
const double EPS = 1e-10;
const double INF = 1e12;
typedef complex<double> Pt;
namespace std {
bool operator<(const Pt& a, const Pt& b) {
  return real(a) != real(b) ? real(a) < real(b) : imag(a) < imag(b);
}
}  // namespace std
double cross(const Pt& a, const Pt& b) { return imag(conj(a) * b); }
double dot(const Pt& a, const Pt& b) { return real(conj(a) * b); }
struct Ln : public vector<Pt> {
  Ln(const Pt& a, const Pt& b) {
    push_back(a);
    push_back(b);
  }
};
int ccw(Pt a, Pt b, Pt c) {
  b -= a;
  c -= a;
  if (cross(b, c) > 0) return +1;
  if (cross(b, c) < 0) return -1;
  if (dot(b, c) < 0) return +2;
  if (norm(b) < norm(c)) return -2;
  return 0;
}
typedef vector<Pt> Pol;
unsigned long long dp[201][201];
Pol pol;
int n;
bool hdp[201][201];
int c = 0;
inline bool hikeru(int a, int b) {
  Pt cn = (pol[a] + pol[b]) / 2.0, pa, pb;
  bool in = false;
  for (int i = (0); (n) > i; i++) {
    pa = pol[i] - cn, pb = pol[(i + 1) % n] - cn;
    if (imag(pa) > imag(pb)) swap(pa, pb);
    if (imag(pa) <= 0 && 0 < imag(pb))
      if (cross(pa, pb) < 0) in = !in;
    if (cross(pa, pb) == 0 && dot(pa, pb) <= 0) {
      return hdp[b][a] = hdp[a][b] = 0;
    }
  }
  if (!in) return hdp[b][a] = hdp[a][b] = 0;
  for (int i = (0); (n) > i; i++) {
    if (pol[i] == pol[a] || pol[i] == pol[b] || pol[(i + 1) % n] == pol[a] ||
        pol[(i + 1) % n] == pol[b])
      continue;
    if ((ccw(pol[i], pol[((i + 1) % n)], pol[a]) *
                 ccw(pol[i], pol[((i + 1) % n)], pol[b]) <=
             0 &&
         ccw(pol[a], pol[b], pol[i]) *
                 ccw(pol[a], pol[b], pol[((i + 1) % n)]) <=
             0))
      return hdp[b][a] = hdp[a][b] = 0;
  }
  return hdp[b][a] = hdp[a][b] = 1;
}
unsigned long long func(int l, int r) {
  if (abs(r - l) == 2) return 1;
  if (dp[l][r]) return dp[l][r];
  unsigned long long ret = 0;
  if (hdp[l][r + 1]) ret += func(l, r + 1);
  ret %= 1000000007;
  if (hdp[l - 1][r]) ret += func(l - 1, r);
  ret %= 1000000007;
  for (int i = (r + 2); (l - 1) > i; i++) {
    if (hdp[i][l] && hdp[i][r]) {
      ret += func(l, i) * func(i, r);
      ret %= 1000000007;
    }
  }
  return dp[l][r] = ret;
}
signed main() {
  scanf("%d", &n);
  pol.resize(n);
  for (int i = (0); (n) > i; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    pol[i] = {(double)x, (double)y};
  }
  for (int i = (0); (n) > i; i++)
    for (int j = (i); (n) > j; j++) hikeru(i, j);
  printf("%d\n", (int)(func(n - 1, 0) % 1000000007));
  return 0;
}
