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 N = 15, mod = 1e9 + 7;
  int n, ans, a[N], dp[(1 << N)], pd[(1 << N)], mv[(1 << N)], sum[(1 << N)],
      Pow[(1 << N)][N];
  long long pow(long long x, long long y, long long mod) {
    return (!y ? 1 : pow(x * x % mod, y / 2, mod) * (y & 1 ? x : 1)) % mod;
  }
  int Cnt(int mask) { return __builtin_popcount(mask); }
  void solve(int mask) {
    int m = 0, res = 0, t = 0;
    mv[mask] = 1;
    for (int i = 0; i < n; i++) {
      if (mask & (1 << i)) {
        t++;
        for (int j = 0; j < i; j++) {
          if (mask & (1 << j)) {
            mv[mask] = 1ll * mv[mask] * (a[i] + a[j]) % mod;
          }
        }
      }
    }
    int inv = (1 << n) - mask - 1;
    m = ((inv + 1) & mask);
    while (m) {
      pd[mask] = (pd[mask] + 1ll * dp[m] * (pd[mask - m] + dp[mask - m]) % mod *
                                 Pow[m][Cnt(mask - m)] % mod) %
                 mod;
      m = (((m | inv) + 1) & mask);
    }
    dp[mask] = (mv[mask] + mod - pd[mask]) % mod;
  }
  int main() {
    mv[0] = 1;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int mask = 0; mask < (1 << n); mask++) {
      Pow[mask][0] = Pow[mask][1] = 1;
      for (int j = 0; j < n; j++)
        if (mask & (1 << j)) Pow[mask][1] = 1ll * Pow[mask][1] * a[j] % mod;
      for (int i = 2; i < N; i++)
        Pow[mask][i] = 1ll * Pow[mask][i - 1] * Pow[mask][1] % mod;
    }
    for (int i = 1; i < (1 << n); i++) {
      solve(i);
    }
    for (int mask = 1; mask < (1 << n); mask++) {
      ans = (ans + 1ll * Cnt(mask) * dp[mask] % mod * Pow[mask][n - Cnt(mask)] %
                       mod * mv[(1 << n) - mask - 1] % mod) %
            mod;
    }
    cout << 1ll * ans * pow(mv[(1 << n) - 1], mod - 2, mod) % mod;
  }
  