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 mod = 1e9 + 7;
  long long a[20], n, f[1 << 14], ans, inv[2000005], now;
  long long dp(int s) {
    if (f[s] != -1) return f[s];
    f[s] = 1;
    for (int t = s & (s - 1); t; t = s & (t - 1)) {
      dp(t), now = 1;
      for (int i = 0; i < n; ++i)
        if (t & (1 << i))
          for (int j = 0; j < n; ++j)
            if ((s & (1 << j)) && !(t & (1 << j)))
              (now *= a[i] * inv[a[i] + a[j]] % mod) %= mod;
      (now *= f[t]) %= mod;
      (f[s] += mod - now) %= mod;
    }
    return f[s];
  }
  int main() {
    scanf("%lld", &n), memset(f, -1, sizeof f), inv[1] = 1;
    for (int i = 2; i <= 2000000; ++i)
      inv[i] = (mod - mod / i) * inv[mod % i] % mod;
    for (int i = 0; i < n; ++i) scanf("%lld", &a[i]);
    f[0] = 0, dp((1 << n) - 1);
    for (int s = 0; s < (1 << n); ++s) {
      int num = __builtin_popcount(s);
      now = 1;
      for (int i = 0; i < n; ++i)
        if (s & (1 << i))
          for (int j = 0; j < n; ++j)
            if (!(s & (1 << j))) (now *= a[i] * inv[a[i] + a[j]] % mod) %= mod;
      (now *= f[s] * num % mod) %= mod;
      (ans += now) %= mod;
    }
    printf("%lld", ans);
    return 0;
  }
  