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;
  long long a, b, c, d, e, i, j, ii, jj, zx, xc, f[29],
      mod = 1000000007LL, pas, G[29][29], kl[17][50009], dp[50009];
  long long xar(long long q, long long w) {
    if (w == 0) return 1LL;
    long long qw = xar(q, w / 2);
    if (w % 2 == 0)
      return (qw * qw) % mod;
    else
      return ((qw * qw) % mod * q) % mod;
  }
  long long ga(long long q, long long w) { return (q * xar(w, mod - 2)) % mod; }
  long long win(long long q, long long w) { return G[q][w]; }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> a;
    for (i = 0; i < a; i++) {
      cin >> f[i];
    }
    for (i = 0; i < a; i++) {
      for (j = 0; j < a; j++) {
        if (i == j) continue;
        G[i][j] = ga(f[i], f[i] + f[j]);
      }
    }
    for (ii = 1; ii < (1 << a); ii++) {
      for (i = 0; i < a; i++) {
        if ((ii & (1 << i)) != 0) continue;
        kl[i][ii] = 1;
        for (j = 0; j < a; j++) {
          if ((ii & (1 << j)) == 0) continue;
          kl[i][ii] *= win(j, i);
          kl[i][ii] %= mod;
        }
      }
    }
    for (ii = 1; ii < (1 << a); ii++) {
      if (__builtin_popcount(ii) == 1) {
        dp[ii] = 0;
        continue;
      }
      for (jj = ii;; jj = ((jj - 1) & ii)) {
        if (jj == 0) break;
        if (jj == ii) continue;
        c = (ii ^ jj);
        zx = 1;
        for (j = 0; j < a; j++) {
          if ((jj & (1 << j)) == 0) continue;
          zx *= kl[j][c];
          zx %= mod;
        }
        dp[ii] += (1 - dp[jj] + mod) * zx;
        dp[ii] %= mod;
      }
    }
    for (ii = 1; ii < (1 << a); ii++) {
      zx = 1;
      for (i = 0; i < a; i++) {
        if ((ii & (1 << i)) == 0) continue;
        for (j = 0; j < a; j++) {
          if ((ii & (1 << j)) != 0) continue;
          zx *= win(i, j);
          zx %= mod;
        }
      }
      c = __builtin_popcount(ii);
      pas += (c * zx) % mod * (1 - dp[ii] + mod);
      pas %= mod;
    }
    cout << pas;
    return 0;
  }
  