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;
int n;
inline int sgn(double xx) {
  if (fabs(xx) < 1e-9) return 0;
  return xx > 0 ? 1 : -1;
}
struct Point {
  double x, y;
  Point() {}
  Point(double X, double Y) : x(X), y(Y) {}
} p[(200 + 50)];
inline Point operator-(const Point &P, const Point &Q) {
  return Point(P.x - Q.x, P.y - Q.y);
}
inline double operator*(const Point &P, const Point &Q) {
  return P.x * Q.y - P.y * Q.x;
}
struct Seg {
  Point p1, p2;
  Seg() {}
  Seg(Point P1, Point P2) : p1(P1), p2(P2) {}
};
inline bool quickquit(Seg s1, Seg s2) {
  double Maxx1 = max(s1.p1.x, s1.p2.x);
  double Maxy1 = max(s1.p1.y, s1.p2.y);
  double Minx1 = min(s1.p1.x, s1.p2.x);
  double Miny1 = min(s1.p1.y, s1.p2.y);
  double Maxx2 = max(s2.p1.x, s2.p2.x);
  double Maxy2 = max(s2.p1.y, s2.p2.y);
  double Minx2 = min(s2.p1.x, s2.p2.x);
  double Miny2 = min(s2.p1.y, s2.p2.y);
  if (sgn(Maxx1 - Minx2) < 0 || sgn(Maxx2 - Minx1) < 0) return 1;
  if (sgn(Maxy1 - Miny2) < 0 || sgn(Maxy2 - Miny1) < 0) return 1;
  return 0;
}
inline bool Cross(Seg s1, Seg s2) {
  if (quickquit(s1, s2)) return 0;
  if (sgn((s1.p1 - s2.p1) * (s2.p2 - s2.p1)) *
          sgn((s1.p2 - s2.p1) * (s2.p2 - s2.p1)) >
      0)
    return 0;
  if (sgn((s2.p1 - s1.p1) * (s1.p2 - s1.p1)) *
          sgn((s2.p2 - s1.p1) * (s1.p2 - s1.p1)) >
      0)
    return 0;
  return 1;
}
bool yes[(200 + 50)][(200 + 50)];
void Init() {
  int i, j, k;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%lf%lf", &p[i].x, &p[i].y);
  for (i = 0; i < n; i++) {
    yes[i][(i + 1) % n] = yes[(i + 1) % n][i] = 1;
    for (j = i + 2; j < n; j++) {
      yes[i][j] = 1;
      for (k = 0; k < n; k++) {
        if (k == i || (k + 1) % n == i) continue;
        if (k == j || (k + 1) % n == j) continue;
        if (Cross(Seg(p[i], p[j]), Seg(p[k], p[(k + 1) % n]))) {
          yes[i][j] = 0;
          break;
        }
      }
      double A1 = 0.0, A2 = 0.0;
      for (k = i; k != j; k = (k + 1) % n)
        A1 += (p[k] - p[i]) * (p[(k + 1) % n] - p[i]);
      for (k = j; k != i; k = (k + 1) % n)
        A2 += (p[k] - p[i]) * (p[(k + 1) % n] - p[i]);
      if (sgn(A1 * A2) <= 0) yes[i][j] = 0;
      yes[j][i] = yes[i][j];
    }
  }
}
long long f[(200 + 50)][(200 + 50)];
void Solve() {
  int i, j, k;
  for (i = 0; i < n; i++) f[i][1] = 1;
  for (j = 2; j < n; j++)
    for (i = 0; i < n; i++) {
      f[i][j] = 0;
      for (k = (i + 1) % n; k != (i + j) % n; k = (k + 1) % n)
        if (yes[i][k] && yes[(i + j) % n][k])
          f[i][j] = (f[i][j] + f[i][(k - i + n) % n] *
                                   f[k][(i + j - k + n) % n] % 1000000007) %
                    1000000007;
    }
  printf("%I64d\n", f[0][n - 1]);
}
int main() {
  Init();
  Solve();
  return 0;
}
