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 P = 1e9 + 7;
struct point {
  point() : x(0), y(0){};
  point(double x, double y) : x(x), y(y){};
  double x;
  double y;
};
point operator-(point a, point b) { return point(a.x - b.x, a.y - b.y); }
bool operator==(point a, point b) { return (a.x == b.x && a.y == b.y); }
vector<point> a;
int n;
double crossProduct(point a, point b) { return (a.x * b.y - a.y * b.x); }
double dotProduct(point a, point b) { return (a.x * b.x + a.y * b.y); }
bool inSeg(point a, point b, point c) {
  return (crossProduct(a - b, a - c) == 0 && dotProduct(a - b, a - c) >= 0 &&
          dotProduct(b - a, b - c) >= 0);
}
bool inLine(point a, point b, point c) {
  return (crossProduct(a - b, a - c) == 0);
}
bool intSeg(point a, point b, point c, point d) {
  if (inSeg(a, b, c) || inSeg(a, b, d) || inSeg(c, d, a) || inSeg(c, d, b))
    return true;
  if (inLine(a, b, c) || inLine(a, b, d) || inLine(a, c, d) || inLine(b, c, d))
    return false;
  if (a == c || b == c || a == d || b == d) return false;
  if (crossProduct(a - b, a - c) > 0 && crossProduct(a - b, a - d) > 0)
    return false;
  if (crossProduct(a - b, a - c) < 0 && crossProduct(a - b, a - d) < 0)
    return false;
  if (crossProduct(c - d, c - a) > 0 && crossProduct(c - d, c - b) > 0)
    return false;
  if (crossProduct(c - d, c - a) < 0 && crossProduct(c - d, c - b) < 0)
    return false;
  return true;
}
bool pointInside(point b) {
  point c(b.x + 1e8 + 1, b.y + 1e8);
  int t = 0;
  for (int i = 0; i < n; i++)
    if (inSeg(a[i], a[(i + 1) % n], b)) return true;
  for (int i = 0; i < n; i++) {
    if (intSeg(a[i], a[(i + 1) % n], b, c)) t++;
  }
  if (t % 2 == 1) return true;
  return false;
}
bool diagInside(int i, int j) {
  point b((a[i].x + a[j].x) / 2, (a[i].y + a[j].y) / 2);
  if (!pointInside(b)) return false;
  for (int k = 0; k < n; k++) {
    if (k == i || k == j) continue;
    if (inSeg(a[i], a[j], a[k])) return false;
  }
  for (int k = 0; k < n; k++) {
    if (k == i || k == j || (k + 1) % n == i || (k + 1) % n == j) continue;
    if (intSeg(a[k], a[(k + 1) % n], a[i], a[j])) return false;
  }
  return true;
}
int main() {
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; i++) cin >> a[i].x >> a[i].y;
  vector<vector<long long> > dp(n);
  for (int i = 0; i < n; i++) {
    dp[i].resize(n);
    if (i < n - 1) dp[i][i + 1] = 1;
  }
  for (int x = 2; x < n; x++) {
    for (int i = 0; i < n - x; i++) {
      dp[i][i + x] = 0;
      if (diagInside(i, i + x)) {
        for (int j = i + 1; j < i + x; j++) {
          dp[i][i + x] = (dp[i][i + x] + dp[i][j] * dp[j][i + x]) % P;
        }
      }
    }
  }
  cout << dp[0][n - 1] << endl;
  return 0;
}
