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 long long mod = 1e9 + 7;
  long long modpow(long long a, long long x) {
    long long tmp = x;
    long long val = a;
    long long ret = 1LL;
    while (tmp > 0) {
      if (tmp % 2 == 1) ret = ret * val % mod;
      val = val * val % mod;
      tmp /= 2;
    }
    return ret;
  }
  long long modinv(long long a) { return modpow(a, mod - 2); }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    vector<vector<long long>> G(1 << n, vector<long long>(n, 1));
    for (int i = 0; i < (1 << n); i++) {
      for (int j = 0; j < n; j++) {
        if (((i >> j) & 1) == 0) continue;
        for (int k = 0; k < n; k++) {
          G[i][k] = G[i][k] * a[j] % mod * modinv(a[j] + a[k]) % mod;
        }
      }
    }
    long long ans = 0LL;
    vector<long long> P(1 << n);
    for (int i = 1; i < (1 << n); i++) {
      long long cnt = __builtin_popcount(i);
      long long P_tmp = 1LL;
      for (int j = i; j > 0; j = ((j - 1) & i)) {
        if (j == i) continue;
        long long G_tmp0 = 1LL;
        for (int k = 0; k < n; k++) {
          if (((j >> k) & 1) == 1) continue;
          if (((i >> k) & 1) == 0) continue;
          G_tmp0 = (G_tmp0 * G[j][k]) % mod;
        }
        P_tmp = (P_tmp - P[j] * G_tmp0 % mod + mod) % mod;
      }
      long long G_tmp1 = 1LL;
      for (int k = 0; k < n; k++) {
        if (((i >> k) & 1) == 1) continue;
        G_tmp1 = (G_tmp1 * G[i][k]) % mod;
      }
      ans = (ans + cnt * P_tmp % mod * G_tmp1) % mod;
      P[i] = P_tmp;
    }
    cout << ans << "\n";
    return 0;
  }
  