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;
typedef struct point {
  double x;
  double y;
  point(const double &x1 = 0, const double &y1 = 0) {
    x = x1;
    y = y1;
  }
} point;
typedef struct line {
  point p1;
  point p2;
  line(const point &s = point(), const point &t = point()) {
    p1 = s;
    p2 = t;
  }
} line;
double eps = 1e-8;
bool operator<(const point &a, const point &b) {
  if (abs(a.x - b.x) > eps) return a.x < b.x;
  return a.y < b.y;
}
bool operator==(const point &a, const point &b) {
  return abs(a.x - b.x) < eps && abs(a.y - b.y) < eps;
}
point operator+(const point &a, const point &b) {
  return point(a.x + b.x, a.y + b.y);
}
point operator-(const point &a, const point &b) {
  return point(a.x - b.x, a.y - b.y);
}
point operator*(const point &a, const double &r) {
  return point(a.x * r, a.y * r);
}
point operator/(const point &a, const double &r) {
  return point(a.x / r, a.y / r);
}
inline double area(const point &p1, const point &p2, const point &p3) {
  point p = p1 - p3;
  point q = p2 - p3;
  return abs(p.x * q.y - p.y * q.x) / 2;
}
enum { LEFT = 1, RIGHT = -1, FRONT = 2, BACK = -2, ON = 0 };
inline int ccw(const point &p1, const point &p2, const point &p3) {
  point p = p2 - p1;
  point q = p3 - p1;
  double r = p.x * q.y - p.y * q.x;
  if (r > eps) {
    return LEFT;
  } else if (r < -eps) {
    return RIGHT;
  } else {
    double t = (p.x * q.x + p.y * q.y) / (p.x * p.x + p.y * p.y);
    if (t > 1) {
      return FRONT;
    } else if (t < 0) {
      return BACK;
    } else {
      return ON;
    }
  }
}
inline bool cross(const line &l1, const line &l2) {
  return ccw(l1.p1, l1.p2, l2.p1) * ccw(l1.p1, l1.p2, l2.p2) <= 0 &&
         ccw(l2.p1, l2.p2, l1.p1) * ccw(l2.p1, l2.p2, l1.p2) <= 0;
}
int n, m = 1000000007;
int f[201][201];
long long dp[201][201][201];
vector<point> v;
int dfs(int x, int y, int z) {
  int i, j;
  long long ans = 0;
  if (dp[x][y][z] != -1) return dp[x][y][z];
  if (y + 1 >= z) return dp[x][y][z] = 1;
  for (i = y; i <= z; i++) {
    if (f[x][i] == 0 || f[y][i] == 0) continue;
    if (ccw(v[x], v[y], v[i]) != LEFT) continue;
    ans += (long long)dfs(i, y, i) * dfs(x, i, z) % m;
    if (ans >= m) ans -= m;
  }
  return dp[x][y][z] = ans;
}
int main() {
  int i, j, k;
  double sum = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    v.push_back(point(x, y));
  }
  for (i = 2; i < n; i++) {
    double s = area(v[0], v[i - 1], v[i]);
    if (ccw(v[0], v[i - 1], v[i]) == RIGHT) s = -s;
    sum += s;
  }
  if (sum < 0) reverse(v.begin(), v.end());
  v.push_back(v[0]);
  for (i = 0; i < v.size(); i++) {
    for (j = 0; j < v.size(); j++) {
      line l = line(v[i], v[j]);
      if (v[i] == v[j]) continue;
      for (k = 0; k < v.size() - 1; k++) {
        if (v[i] == v[k] || v[j] == v[k]) continue;
        if (v[i] == v[k + 1] || v[j] == v[k + 1]) continue;
        if (cross(l, line(v[k], v[k + 1]))) break;
      }
      if (k == v.size() - 1) f[i][j] = 1;
    }
  }
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= n; j++) {
      for (k = 0; k <= n; k++) {
        dp[i][j][k] = -1;
      }
    }
  }
  printf("%d", dfs(0, 1, n));
  printf("\n");
  return 0;
}
