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 MX = 200 + 23;
const long long MOD = 1e9 + 7;
struct VECTOR {
  long long x, y;
  VECTOR(long long X = 0, long long Y = 0) { x = X, y = Y; }
  VECTOR operator+(const VECTOR &B) const { return VECTOR(x + B.x, y + B.y); }
  VECTOR operator-(const VECTOR &B) const { return VECTOR(x - B.x, y - B.y); }
  long long operator*(const VECTOR &B) const { return x * B.y - y * B.x; }
  bool operator==(const VECTOR &B) const { return x == B.x && y == B.y; }
  void Read() { cin >> x >> y; }
} P[MX];
struct SEG {
  VECTOR st, d;
  SEG() { st = d = VECTOR(0, 0); }
  SEG(VECTOR __st, VECTOR __ed) { st = __st, d = __ed - __st; };
  void turn() {
    st = st + d;
    d = VECTOR(0, 0) - d;
  }
};
int sgn(long long x) {
  if (!x) return 0;
  return x > 0 ? 1 : -1;
}
bool PingXing(VECTOR A, VECTOR B) { return A.x * B.y - A.y * B.x == 0; }
bool online(SEG a, VECTOR b) {
  bool xless = b.x < a.st.x && b.x < (a.st + a.d).x;
  bool xmore = b.x > a.st.x && b.x > (a.st + a.d).x;
  bool yless = b.y < a.st.y && b.y < (a.st + a.d).y;
  bool ymore = b.y > a.st.y && b.y > (a.st + a.d).y;
  return (xless + xmore + yless + ymore) == 0;
}
int ok(SEG a, SEG b) {
  long long test1 = sgn(a.d * (b.st - a.st)) * sgn(a.d * (b.st + b.d - a.st));
  long long test2 = sgn(b.d * (a.st - b.st)) * sgn(b.d * (a.st + a.d - b.st));
  if (test1 > 0 || test2 > 0) {
    return 2;
  }
  if (PingXing(a.d, b.d)) {
    if (online(a, b.st) || online(a, b.st + b.d) || online(b, a.st) ||
        online(b, a.st + a.d)) {
      if (sgn(a.d.x) != sgn(b.d.x) || sgn(a.d.y) != sgn(b.d.y)) {
        b.turn();
      }
      if (a.st + a.d == b.st || b.st + b.d == a.st) return 1;
      return 0;
    }
    return 2;
  }
  return (test1 <= 0 && test2 <= 0);
}
long long fad;
int n;
int check(int l, int r) {
  SEG Main(P[l], P[r]);
  long long S = fad, s1 = 0, s2 = 0;
  for (int i = l; i <= r; ++i) {
    if (i == r) {
      s1 += P[r] * P[l];
    } else {
      s1 += P[i] * P[i + 1];
    }
  }
  for (int i = r;; i = (i + 1) % n) {
    if (i == l) {
      s2 += P[l] * P[r];
      break;
    }
    s2 += P[i] * P[(i + 1) % n];
  }
  if (abs(S) != abs(s1) + abs(s2)) return 0;
  for (int i = 0; i < n; ++i) {
    SEG t(P[i], P[(i + 1) % n]);
    int judge = ok(Main, t);
    if (i == r && ((i + 1) % n == l)) continue;
    if (judge == 0) return 0;
    if (judge == 1) {
      if (i == l || (i + 1) % n == l || i == r || (i + 1) % n == r) {
        continue;
      }
      return 0;
    }
  }
  return 1;
}
long long dp[MX][MX];
void test() {
  int n;
  cin >> n;
  while (n--) {
    int x, y;
    cin >> x >> y;
    printf("(%d,%d)\n", x, y);
  }
  SEG a(VECTOR(1, 0), VECTOR(-1, 0)), b(VECTOR(0, 0), VECTOR(1, 0));
  cout << ok(a, b) << endl;
  return;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    P[i].Read();
  }
  for (int i = 0; i < n; ++i) {
    fad += P[i] * P[(i + 1) % n];
  }
  for (int i = 1; i < n; ++i) {
    dp[i - 1][i] = 1;
  }
  for (int len = 2; len <= n; ++len) {
    for (int st = 0; st + len < n; ++st) {
      int ed = st + len;
      if (!check(st, ed)) {
        continue;
      }
      for (int k = st + 1; k < ed; ++k) {
        dp[st][ed] = (dp[st][ed] + dp[st][k] * dp[k][ed]) % MOD;
      }
    }
  }
  cout << dp[0][n - 1] << endl;
  return 0;
}
