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 N = 2e2 + 10;
const double eps = 1e-10;
const long long mod = 1e9 + 7;
int dcmp(double x) {
  if (x > eps) return 1;
  if (x < -eps) return -1;
  return 0;
}
class Point {
 public:
  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 a) { return Point(a * x, a * y); }
  Point operator/(double a) { return Point(x / a, y / a); }
  double abs() { return sqrt(norm()); }
  double norm() { return x * x + y * y; }
  bool operator<(const Point &p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const Point &p) const {
    return dcmp(x - p.x) == 0 && dcmp(y - p.y) == 0;
  }
  void read() { scanf("%lf%lf", &x, &y); }
} p[N];
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; }
static const int COUNTER_CLOCKWISE = 1;
static const int CLOCKWISE = -1;
static const int ONLINE_BACK = 2;
static const int ONLINE_FRONT = -2;
static const int ON_SEGMENT = 0;
int ccw(Point p0, Point p1, Point p2) {
  Point a = p1 - p0;
  Point b = p2 - p0;
  if (cross(a, b) > eps) return COUNTER_CLOCKWISE;
  if (cross(a, b) < -eps) return CLOCKWISE;
  if (dot(a, b) < -eps) return ONLINE_BACK;
  if (a.norm() < b.norm()) return ONLINE_FRONT;
  return ON_SEGMENT;
}
bool LineLineIntersect(Point p1, Point p2, Point p3, Point p4) {
  return (ccw(p1, p2, p3) * ccw(p1, p2, p4) <= 0 &&
          ccw(p3, p4, p1) * ccw(p3, p4, p2) <= 0);
}
vector<Point> g;
bool LinePolygonIntersect(Point p1, Point p2, vector<Point> p) {
  int n = p.size();
  for (int i = 0; i < n; ++i) {
    if (LineLineIntersect(p1, p2, p[i], p[(i + 1) % n])) return true;
  }
  return false;
}
long long dp[N][N];
void cal(int l, int r) {
  if (dp[l][r] >= 0) return;
  if (l == r - 1) {
    dp[l][r] = 1;
    return;
  }
  dp[l][r] = 0;
  for (int i = l + 1; i < r; i++) {
    if (cross((p[l] - p[r]), (p[i] - p[r])) > 0) {
      cal(l, i);
      cal(i, r);
      (dp[l][r] += dp[l][i] * dp[i][r]) %= mod;
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    p[i].read();
    g.push_back(p[i]);
  }
  long long s = 0;
  for (int i = 0; i < n; i++) s += cross(p[i], p[(i + 1) % n]);
  if (s < 0) reverse(p, p + n);
  memset(dp, -1, sizeof(dp));
  cal(0, n - 1);
  printf("%lld\n", dp[0][n - 1]);
  return 0;
}
