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 MAXN = 205;
const long long modulo = 1000000007;
struct point {
  long long x, y;
};
point operator-(point A, point B) {
  point C;
  C.x = A.x - B.x;
  C.y = A.y - B.y;
  return C;
}
bool operator==(point A, point B) { return (A.x == B.x && A.y == B.y); }
long long cross(point A, point B) { return (A.x * B.y - A.y * B.x); }
long long dot(point A, point B) { return (A.x * B.x + A.y * B.y); }
struct Line {
  long long A, B, C;
  Line(point P, point R) {
    A = R.y - P.y;
    B = P.x - R.x;
    C = (P.y * R.x) - (P.x * R.y);
  }
};
long long sgn(long long x) {
  if (x > 0) return 1;
  if (x < 0) return -1;
  if (x == 0) return 0;
}
long long f(Line L, point P) { return sgn(L.A * P.x + L.B * P.y + L.C); }
bool OnSegment(point A, point B, point X) {
  if (f(Line(A, B), X) != 0) return false;
  if (dot(A - X, B - X) < 0)
    return true;
  else
    return false;
}
bool intersect(point A, point B, point C, point D) {
  if (A == C || A == D || B == C || B == D) return false;
  if (f(Line(A, B), C) * f(Line(A, B), D) >= 0) return false;
  if (f(Line(C, D), A) * f(Line(C, D), B) >= 0) return false;
  return true;
}
point a[MAXN];
long long dp[MAXN][MAXN];
int main() {
  int N;
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> a[i].x >> a[i].y;
  }
  a[N + 1] = a[1];
  long long Area = 0;
  for (int i = 1; i <= N; i++) {
    Area += cross(a[i], a[i + 1]);
  }
  Area = abs(Area);
  for (int d = 1; d < N - 1; d++) {
    for (int l = 1; l + d <= N; l++) {
      int r = l + d;
      if (d == 1) {
        dp[l][r] = 1;
        continue;
      } else
        dp[l][r] = 0;
      bool bad = false;
      for (int i = 1; i <= N; i++) {
        if (i != l && i != r && OnSegment(a[l], a[r], a[i])) bad = true;
      }
      if (bad) continue;
      for (int i = 1; i <= N; i++) {
        int next = i + 1;
        if (next == N + 1) next = 1;
        if (intersect(a[l], a[r], a[i], a[next])) bad = true;
      }
      if (bad) continue;
      long long AreaX = 0, AreaY = 0;
      for (int i = l; i <= r; i++) {
        int next = i + 1;
        if (next == r + 1) next = l;
        AreaX += cross(a[i], a[next]);
      }
      int j = 1;
      while (j <= N) {
        int next;
        if (j == l)
          next = r;
        else
          next = j + 1;
        if (next == N + 1) next = 1;
        AreaY += cross(a[j], a[next]);
        j = next;
        if (j == 1) j = N + 1;
      }
      if (Area != abs(AreaX) + abs(AreaY)) continue;
      for (int i = l + 1; i < r; i++) {
        dp[l][r] += (dp[l][i] * dp[i][r]) % modulo;
        dp[l][r] = dp[l][r] % modulo;
      }
    }
  }
  dp[1][N] = 0;
  for (int i = 2; i < N; i++) {
    dp[1][N] += (dp[1][i] * dp[i][N]) % modulo;
    dp[1][N] = dp[1][N] % modulo;
  }
  cout << dp[1][N] << endl;
  return 0;
}
