description: <image>
  
  William is not only interested in trading but also in betting on sports matches. n teams participate in each match. Each team is characterized by strength a_i. Each two teams i < j play with each other exactly once. Team i wins with probability (a_i)/(a_i + a_j) and team j wins with probability (a_j)/(a_i + a_j).
  
  The team is called a winner if it directly or indirectly defeated all other teams. Team a defeated (directly or indirectly) team b if there is a sequence of teams c_1, c_2, ... c_k such that c_1 = a, c_k = b and team c_i defeated team c_{i + 1} for all i from 1 to k - 1. Note that it is possible that team a defeated team b and in the same time team b defeated team a.
  
  William wants you to find the expected value of the number of winners.
  
  Input
  
  The first line contains a single integer n (1 ≤ n ≤ 14), which is the total number of teams participating in a match.
  
  The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6) — the strengths of teams participating in a match.
  
  Output
  
  Output a single integer — the expected value of the number of winners of the tournament modulo 10^9 + 7.
  
  Formally, let M = 10^9+7. It can be demonstrated that the answer can be presented as a irreducible fraction p/q, where p and q are integers and q not ≡ 0 \pmod{M}. Output a single integer equal to p ⋅ q^{-1} mod M. In other words, output an integer x such that 0 ≤ x < M and x ⋅ q ≡ p \pmod{M}.
  
  Examples
  
  Input
  
  
  2
  1 2
  
  
  Output
  
  
  1
  
  
  Input
  
  
  5
  1 5 2 11 14
  
  
  Output
  
  
  642377629
  
  Note
  
  To better understand in which situation several winners are possible let's examine the second test:
  
  One possible result of the tournament is as follows (a → b means that a defeated b):
  
    * 1 → 2 
    * 2 → 3 
    * 3 → 1 
    * 1 → 4 
    * 1 → 5 
    * 2 → 4 
    * 2 → 5 
    * 3 → 4 
    * 3 → 5 
    * 4 → 5 
  
  
  
  Or more clearly in the picture:
  
  <image>
  
  In this case every team from the set \{ 1, 2, 3 \} directly or indirectly defeated everyone. I.e.:
  
    * 1st defeated everyone because they can get to everyone else in the following way 1 → 2, 1 → 2 → 3, 1 → 4, 1 → 5. 
    * 2nd defeated everyone because they can get to everyone else in the following way 2 → 3, 2 → 3 → 1, 2 → 4, 2 → 5. 
    * 3rd defeated everyone because they can get to everyone else in the following way 3 → 1, 3 → 1 → 2, 3 → 4, 3 → 5. 
  
  
  
  Therefore the total number of winners is 3.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int MX = (1 << 15) + 2, md = 1000000007;
  int n, m, i, j, k, msk, a[15], b[MX], pos[15];
  long long rp, rq, zp, zq, curp, curq, cp[MX], cq[MX], p[15][15], q[15][15];
  long long pw(long long a, int b) {
    if (b == 0) return 1LL;
    if (b & 1) return (pw(a, b - 1) * a) % md;
    long long x = pw(a, b / 2);
    return (x * x) % md;
  }
  void mul(long long& xp, long long& xq, long long yp, long long yq) {
    xp = (xp * yp) % md;
    xq = (xq * yq) % md;
  }
  void sub(long long& xp, long long& xq, long long yp, long long yq) {
    xp = (xp * yq + md - ((xq * yp) % md)) % md;
    xq = (xq * yq) % md;
  }
  void inc(long long& xp, long long& xq, long long yp, long long yq) {
    xp = (xp * yq + xq * yp) % md;
    xq = (xq * yq) % md;
  }
  int main() {
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      for (j = 0; j < i; j++) {
        p[i][j] = a[i];
        p[j][i] = a[j];
        q[i][j] = q[j][i] = a[i] + a[j];
      }
    }
    rp = 0;
    rq = 1;
    for (i = 1; i < (1 << n); i++) {
      b[i] = b[i / 2] + (i & 1);
      cp[i] = zp = 1;
      cq[i] = zq = 1;
      for (m = j = 0; j < n; j++)
        if ((i >> j) & 1) {
          pos[m++] = j;
          for (k = 0; k < n; k++)
            if (((i >> k) & 1) == 0) mul(zp, zq, p[j][k], q[j][k]);
        }
      for (msk = (i & (i - 1)); msk > 0; msk = ((msk - 1) & i)) {
        curp = cp[msk];
        curq = cq[msk];
        for (j = 0; j < m; j++)
          if ((msk >> pos[j]) & 1)
            for (k = 0; k < m; k++)
              if (((msk >> pos[k]) & 1) == 0)
                mul(curp, curq, p[pos[j]][pos[k]], q[pos[j]][pos[k]]);
        sub(cp[i], cq[i], curp, curq);
      }
      inc(rp, rq, (((b[i] * cp[i]) % md) * zp) % md, (cq[i] * zq) % md);
    }
    printf("%lld\n", (rp * pw(rq, md - 2)) % md);
    return 0;
  }
  