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 = 200;
struct pt {
  long long x, y;
  pt() {}
  pt(long long x, long long y) : x(x), y(y) {}
  pt operator-(const pt &r) const { return pt(x - r.x, y - r.y); }
  pt operator+(const pt &r) const { return pt(x + r.x, y + r.y); }
  friend long long scal(const pt &l, const pt &r) {
    return l.x * r.x + l.y * r.y;
  }
  friend long long vect(const pt &l, const pt &r) {
    return l.x * r.y - l.y * r.x;
  }
  long double len() const { return sqrtl(scal(*this, *this)); }
};
pt a[N];
int n;
const long long mod = 1000000007;
long long dp[N][N];
long double eps = 1e-9;
inline long long area(pt a, pt b, pt c) {
  return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}
inline bool intersect_1(long long a, long long b, long long c, long long d) {
  if (a > b) swap(a, b);
  if (c > d) swap(c, d);
  return max(a, c) <= min(b, d);
}
bool check1(pt a, pt b, pt c, pt d) {
  if (intersect_1(a.x, b.x, c.x, d.x) && intersect_1(a.y, b.y, c.y, d.y)) {
    if ((area(a, b, c) <= 0 && area(a, b, d) >= 0) ||
        (area(a, b, c) >= 0 && area(a, b, d) <= 0)) {
      if ((area(c, d, a) <= 0 && area(c, d, b) >= 0) ||
          (area(c, d, a) >= 0 && area(c, d, b) <= 0)) {
        return true;
      }
    }
  }
  return false;
}
bool check2(pt a, pt b, pt c) {
  if (vect(a - b, c - b) == 0) {
    if (c.x >= min(a.x, b.x) && c.x <= max(a.x, b.x) && c.y >= min(a.y, b.y) &&
        c.y <= max(a.y, b.y)) {
      return true;
    }
  }
  return false;
}
long long solve(int l, int r) {
  if (r <= l + 1) {
    return dp[l][r] = 1;
  }
  if (dp[l][r] != -1) {
    return dp[l][r];
  }
  long long sq = 0;
  for (int i = l; i <= r; ++i) {
    int j = i + 1;
    if (j > r) j = l;
    sq += vect(a[i], a[j]);
  }
  if (sq < 0) {
    return dp[l][r] = 0;
  }
  sq = 0;
  for (int i = r;; i = (i + 1) % n) {
    int j = (i + 1) % n;
    if (i == l) {
      j = r;
    }
    sq += vect(a[i], a[j]);
    if (i == l) {
      break;
    }
  }
  if (sq < 0) {
    return dp[l][r] = 0;
  }
  bool ok = true;
  for (int i = 0; i < n && ok; ++i) {
    int j = (i + 1) % n;
    if (i == l || j == l) {
      continue;
    }
    if (i == r || j == r) {
      continue;
    }
    if (check1(a[r], a[l], a[i], a[j])) {
      ok = false;
    }
  }
  if (r - l != 1) {
    if (check2(a[r], a[l], a[l + 1])) {
      ok = false;
    }
    if (check2(a[r], a[l], a[r - 1])) {
      ok = false;
    }
  }
  if (r - l != n - 1) {
    if (check2(a[r], a[l], a[(l - 1 + n) % n])) {
      ok = false;
    }
    if (check2(a[r], a[l], a[(r + 1) % n])) {
      ok = false;
    }
  }
  if (!ok) {
    return dp[l][r] = 0;
  }
  dp[l][r] = 0;
  for (int i = l + 1; i < r; ++i) {
    dp[l][r] += solve(l, i) * solve(i, r) % mod;
    if (dp[l][r] >= mod) {
      dp[l][r] -= mod;
    }
  }
  return dp[l][r];
}
int main() {
  ios_base::sync_with_stdio(false);
  memset(dp, 255, sizeof(dp));
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i].x >> a[i].y;
  }
  long long square = 0;
  for (int i = 0; i < n; ++i) {
    int j = (i + 1) % n;
    square += vect(a[i], a[j]);
  }
  if (square < 0) {
    reverse(a, a + n);
  }
  cout << solve(0, n - 1) << endl;
  return 0;
}
