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;
double INF = 1e100, EPS = 1e-12;
struct PT {
  double x, y;
  PT() {}
  PT(double x, double y) : x(x), y(y) {}
  PT(const PT &p) : x(p.x), y(p.y) {}
  PT operator+(const PT &p) const { return PT(x + p.x, y + p.y); }
  PT operator-(const PT &p) const { return PT(x - p.x, y - p.y); }
  PT operator*(double c) const { return PT(x * c, y * c); }
  PT operator/(double c) const { return PT(x / c, y / c); }
};
double dot(PT p, PT q) { return p.x * q.x + p.y * q.y; }
double dist2(PT p, PT q) { return dot(p - q, p - q); }
double cross(PT p, PT q) { return p.x * q.y - p.y * q.x; }
PT ProjectPointLine(PT a, PT b, PT c) {
  return a + (b - a) * dot(c - a, b - a) / dot(b - a, b - a);
}
PT ProjectPointSegment(PT a, PT b, PT c) {
  double r = dot(b - a, b - a);
  if (fabs(r) < EPS) return a;
  r = dot(c - a, b - a) / r;
  if (r < 0) return a;
  if (r > 1) return b;
  return a + (b - a) * r;
}
double DistancePointSegment(PT a, PT b, PT c) {
  return sqrt(dist2(c, ProjectPointSegment(a, b, c)));
}
bool PointOnSegment(PT a, PT b, PT c) {
  if (dist2(ProjectPointSegment(a, b, c), c) < EPS) return true;
  return false;
}
bool LinesParallel(PT a, PT b, PT c, PT d) {
  return fabs(cross(b - a, c - d)) < EPS;
}
bool LinesCollinear(PT a, PT b, PT c, PT d) {
  return LinesParallel(a, b, c, d) && fabs(cross(a - b, a - c)) < EPS &&
         fabs(cross(c - d, c - a)) < EPS;
}
bool SegmentsIntersect(PT a, PT b, PT c, PT d) {
  if (LinesCollinear(a, b, c, d)) {
    if (dist2(a, c) < EPS || dist2(a, d) < EPS || dist2(b, c) < EPS ||
        dist2(b, d) < EPS)
      return true;
    if (dot(c - a, c - b) > 0 && dot(d - a, d - b) > 0 && dot(c - b, d - b) > 0)
      return false;
    return true;
  }
  if (cross(d - a, b - a) * cross(c - a, b - a) > 0) return false;
  if (cross(a - c, d - c) * cross(b - c, d - c) > 0) return false;
  return true;
}
PT ComputeLineIntersection(PT a, PT b, PT c, PT d) {
  b = b - a;
  d = c - d;
  c = c - a;
  return a + b * cross(c, d) / cross(b, d);
}
bool PointInPolygon(const vector<PT> &p, PT q) {
  bool c = 0;
  for (int i = 0; i < p.size(); i++) {
    int j = (i + 1) % p.size();
    if ((p[i].y <= q.y && q.y < p[j].y || p[j].y <= q.y && q.y < p[i].y) &&
        q.x < p[i].x + (p[j].x - p[i].x) * (q.y - p[i].y) / (p[j].y - p[i].y))
      c = !c;
  }
  return c;
}
int n;
vector<PT> polygon;
long long dp[205][205];
long long solve(int i, int j) {
  if (dp[i][j] >= 0LL) return dp[i][j];
  if (j == i + 1) return dp[i][j] = 1LL;
  dp[i][j] = 0LL;
  for (int k = 0; k < n; k++) {
    if (k == i || k == j) continue;
    if (PointOnSegment(polygon[i], polygon[j], polygon[k])) {
      return 0LL;
    }
  }
  for (int k = 0; k < n; k++) {
    if (k == j || k == j - 1 || k == i || k == i - 1 ||
        (i == 0 && k == n - 1) || (j == 0 && k == n - 1))
      continue;
    if (SegmentsIntersect(polygon[i], polygon[j], polygon[k],
                          polygon[(k + 1) % n])) {
      return 0LL;
    }
  }
  if (i != (j + 1) % n && j != (i + 1) % n &&
      !PointInPolygon(polygon, (polygon[i] + polygon[j]) / 2)) {
    return 0LL;
  }
  for (int k = i + 1; k < j; k++) {
    dp[i][j] += (solve(i, k) * solve(k, j)) % 1000000007;
    dp[i][j] %= 1000000007;
  }
  return dp[i][j];
}
int main() {
  scanf(" %d", &n);
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf(" %d %d", &x, &y);
    polygon.push_back(PT(x, y));
  }
  for (int i = 0; i < 205; i++)
    for (int j = 0; j < 205; j++) dp[i][j] = -1;
  printf("%lld\n", solve(0, n - 1));
  return 0;
}
