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;
struct Point {
  double x, y;
  Point(double _x = 0, double _y = 0) {
    x = _x;
    y = _y;
  }
  Point operator+(Point p) { return Point(x + p.x, y + p.y); }
  Point operator-(Point p) { return Point(x - p.x, y - p.y); }
  Point operator*(double t) { return Point(x * t, y * t); }
  Point operator/(double t) { return Point(x / t, y / t); }
};
double dot(Point a, Point b) { return a.x * b.x + a.y * b.y; }
double cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
double dist(Point u, Point v) {
  return sqrt((u.x - v.x) * (u.x - v.x) + (u.y - v.y) * (u.y - v.y));
}
double norm(Point p) { return hypot(p.x, p.y); }
double arg(Point p) { return atan2(p.y, p.x); }
Point polar(double len, double al) {
  return Point(len * cos(al), len * sin(al));
}
bool intersect(Point a, Point b, Point c, Point d) {
  return cross(a - b, c - b) * cross(a - b, d - b) < 1e-8 &&
         cross(c - d, a - d) * cross(c - d, b - d) < 1e-8 &&
         min(a.x, b.x) < max(c.x, d.x) + 1e-8 &&
         min(c.x, d.x) < max(a.x, b.x) + 1e-8 &&
         min(a.y, b.y) < max(c.y, d.y) + 1e-8 &&
         min(c.y, d.y) < max(a.y, b.y) + 1e-8;
}
bool onseg(Point a, Point p, Point q) {
  if (fabs(cross(p - a, q - a)) > 1e-8) return false;
  return dot(p - a, q - a) < 1e-8;
}
int inpoly(Point *p, int n, Point a) {
  double rot = 0;
  for (int i = 0; i < n; i++)
    if (onseg(a, p[i], p[i + 1])) return 0;
  for (int i = 0; i < n; i++) {
    double al =
        dot(p[i] - a, p[i + 1] - a) / (dist(p[i], a) * dist(p[i + 1], a));
    al = min(al, 1.0);
    al = max(al, -1.0);
    al = acos(al);
    if (cross(p[i] - a, p[i + 1] - a) > 0)
      rot += al;
    else
      rot -= al;
  }
  rot = fabs(rot);
  if (rot > M_PI * 3 / 2)
    return 1;
  else if (rot > M_PI / 2)
    return 0;
  else
    return -1;
}
int N;
Point p[220];
int can[220][220], dp[220][220], vst[220][220];
int calc(int s, int e) {
  if (vst[s][e]) return dp[s][e];
  if (e <= s + 1) return 1;
  for (int i = s + 1; i < e; i++) {
    if (!can[s][i] || !can[i][e]) continue;
    dp[s][e] = (dp[s][e] + (long long)calc(s, i) * calc(i, e) % 1000000007) %
               1000000007;
  }
  vst[s][e] = 1;
  return dp[s][e];
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%lf%lf", &p[i].x, &p[i].y);
  p[N] = p[0];
  for (int i = 0; i < N; i++)
    for (int j = i + 2; j < N; j++) {
      if (inpoly(p, N, (p[i] + p[j]) / 2) != 1) continue;
      int k;
      for (k = 0; k < N; k++) {
        if (k == i || k == j || (k + 1) % N == i || (k + 1) % N == j) continue;
        if (intersect(p[i], p[j], p[k], p[k + 1])) break;
      }
      if (k == N) can[i][j] = 1;
    }
  for (int i = 0; i < N; i++) can[i][i + 1] = 1;
  printf("%d\n", calc(0, N - 1));
  return 0;
}
