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 + 10, MN = 205, MOD = 1e9 + 7;
const long double Pi = 3.14159265, eps = 1e-9;
int n, dp[MN][MN];
bool is[MN][MN];
bool eq(long double a, long double b) { return a - eps <= b && b <= a + eps; }
struct point {
  long long x, y;
  point(long long x = 0, long long y = 0) : x(x), y(y) {}
  void in() { cin >> x >> y; }
  point operator-(const point &a) { return {x - a.x, y - a.y}; }
  point operator+(const point &a) { return {x + a.x, y + a.y}; }
  long long operator^(const point &a) { return a.y * x - a.x * y; }
  long long len2() { return x * x + y * y; }
  long long operator*(const point &a) { return x * a.x + y * a.y; }
  long double angle() {
    return atan2(y, x) < 0 ? atan2(y, x) + Pi + Pi : atan2(y, x);
  }
} p[MN];
struct line {
  point a, ab;
  line() : a(), ab() {}
  line(point a, point b) : a(a), ab(b - a) {}
  point b() { return a + ab; }
};
bool sig(line L, point p) { return ((p - L.a) ^ L.ab) > 0; }
bool ison(line L, point p) {
  if (((p - L.a) ^ L.ab) != 0) return false;
  long long X = ((p - L.a) * L.ab);
  if (X < 0 || X > L.ab.len2()) return false;
  return true;
}
bool check(line L1, line L2) {
  if (ison(L1, L2.a) || ison(L1, L2.b()) || ison(L2, L1.a) || ison(L2, L1.b()))
    return true;
  return (sig(L1, L2.a) != sig(L1, L2.b())) &&
         (sig(L2, L1.a) != sig(L2, L1.b()));
}
long double D(long double a, long double b) {
  if (b > a) return b - a;
  return Pi + Pi - (a - b);
}
bool chck2(point A, point B, point C) {
  long double a = A.angle(), b = B.angle(), c = C.angle();
  return eq(D(a, c), D(a, b) + D(b, c));
}
void ok(int &x) {
  if (x >= MOD) x -= MOD;
}
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    p[i].in();
  }
  long long area = 0;
  for (int i = 0; i < n; ++i) area += p[i] ^ p[(i + 1) % n];
  if (area < 0) reverse(p, p + n);
  for (int i = 0; i < n; ++i) {
    dp[i][(i + 1) % n] = 1;
    is[i][(i + 1) % n] = 1;
    is[(i + 1) % n][i] = 1;
  }
  for (int i = 0; i < n; ++i)
    for (int j = (i + 2) % n; j != (i + n - 1) % n; (++j) %= n) {
      bool Fail = false;
      for (int l = 0; l < n; ++l)
        if (l != i && l != j && ison(line(p[i], p[j]), p[l])) Fail = true;
      for (int l = 0; l < n; ++l)
        if (l != i && (l + 1) % n != i && l != j && (l + 1) % n != j &&
            check(line(p[i], p[j]), line(p[l], p[(l + 1) % n])))
          Fail = true;
      if (Fail) continue;
      if (!chck2((p[(i + 1) % n] - p[i]), (p[j] - p[i]),
                 (p[(i + n - 1) % n] - p[i])))
        continue;
      if (!chck2((p[(j + 1) % n] - p[j]), (p[i] - p[j]),
                 (p[(j + n - 1) % n] - p[j])))
        continue;
      is[i][j] = true;
    }
  for (int sz = 2; sz < n; ++sz)
    for (int l = 0; l < n; ++l) {
      int r = (l + sz) % n;
      if (!is[l][r]) continue;
      for (int an = (l + 1) % n; an != r; an = (an + 1) % n)
        if (is[an][l] && is[an][r]) {
          dp[l][r] += 1ll * dp[l][an] * dp[an][r] % MOD;
          ok(dp[l][r]);
        }
    }
  cout << dp[0][n - 1] << '\n';
  return 0;
}
