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 long long P = 1000000007;
int n;
struct Point {
  long long x, y;
  Point operator-(const Point &b) const {
    Point c;
    c.x = x - b.x;
    c.y = y - b.y;
    return c;
  }
  Point operator+(const Point &b) const {
    Point c;
    c.x = x + b.x;
    c.y = y + b.y;
    return c;
  }
  Point operator*(const double xx) const {
    Point c;
    c.x = x * xx;
    c.y = y * xx;
    return c;
  }
  long long operator*(const Point &b) const { return x * b.y - b.x * y; }
  long long operator&(const Point &b) const { return x * b.x + b.y * y; }
} po[200 + 10];
long long dp[200 + 10][200 + 10];
struct Line {
  Point a, b;
};
Point getIntersection(Line l1, Line l2) {
  double t = ((l1.b - l1.a) * (l2.a - l1.a)) /
             ((l1.b - l1.a) * (l2.a - l1.a) + (l2.b - l1.a) * (l1.b - l1.a));
  return l2.a + (l2.b - l2.a) * t;
}
bool cross(Line l1, Line l2) {
  if ((l1.b - l1.a) * (l2.b - l2.a) == 0) {
    if ((l2.a - l1.a) * (l1.b - l1.a) != 0) return false;
    if (((l1.a - l2.a) & (l1.b - l2.a)) <= 0 ||
        ((l1.a - l2.b) & (l1.b - l2.b)) <= 0)
      return true;
    if (((l2.a - l1.a) & (l2.b - l1.a)) <= 0 ||
        ((l2.a - l1.b) & (l2.b - l1.b)) <= 0)
      return true;
    return false;
  }
  Point pp = getIntersection(l1, l2);
  if (((l1.a - pp) & (l1.b - pp)) <= 0 && ((l2.a - pp) & (l2.b - pp)) <= 0)
    return true;
  else
    return false;
}
long long dfs(int s, int t) {
  if (t - s <= 1) return 1;
  if (dp[s][t] != -1) return dp[s][t];
  dp[s][t] = 0;
  for (int i = s + 1; i < t; i++) {
    if ((po[i] - po[s]) * (po[t] - po[s]) > 0)
      dp[s][t] = (dp[s][t] + (dfs(s, i) * dfs(i, t)) % P) % P;
  }
  return dp[s][t];
}
void check() {
  long long area = 0;
  for (int i = 2; i < n; i++) {
    area += (po[i] - po[1]) * (po[i + 1] - po[1]);
  }
  if (area < 0) reverse(po + 1, po + n + 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d%I64d", &po[i].x, &po[i].y);
  }
  check();
  memset(dp, -1, sizeof(dp));
  cout << dfs(1, n) << endl;
  return 0;
}
