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);
  const long long MAX_TEAMS = (14);
  const long long MAX_MASK = (5e6);
  long long Powers[MAX_TEAMS + 1];
  long long Strenght[MAX_TEAMS];
  long long Proba[MAX_TEAMS][MAX_TEAMS];
  long long ProbaMask[MAX_MASK];
  long long Transpo[1 << MAX_TEAMS];
  long long DP[1 << MAX_TEAMS];
  long long nbTeams;
  long long GetBit(long long mask, long long index) {
    return (mask / Powers[index]) % 3;
  }
  long long fast_expo(long long a, long long n) {
    long long ans = 1, cur = a;
    for (; n > 0; n >>= 1, cur = (cur * cur) % MOD) {
      if (n & 1) ans = (ans * cur) % MOD;
    }
    return ans;
  }
  void Solve() {
    scanf("%lld", &nbTeams);
    Powers[0] = 1;
    for (long long i = 0; i < nbTeams; i++) {
      scanf("%lld", &Strenght[i]);
      Powers[i + 1] = 3 * Powers[i];
    }
    long long nbMasks = Powers[nbTeams];
    for (long long i = 0; i < nbTeams; i++) {
      for (long long j = 0; j < nbTeams; j++) {
        Proba[i][j] = Strenght[i] * fast_expo(Strenght[i] + Strenght[j], MOD - 2);
        Proba[i][j] %= MOD;
      }
    }
    for (long long m = 0; m < (1 << nbTeams); m++) {
      for (long long j = 0; j < nbTeams; j++) {
        if ((m >> j) & 1) Transpo[m] += Powers[j];
      }
    }
    ProbaMask[nbMasks - 1] = 1;
    for (long long m = nbMasks - 2; m >= 0; m--) {
      long long in = 0;
      for (long long i = 0; i < nbTeams; i++) {
        if (GetBit(m, i) < 2) in = i;
      }
      ProbaMask[m] = ProbaMask[m + (2 - GetBit(m, in)) * Powers[in]];
      for (long long i = 0; i < nbTeams; i++) {
        if (GetBit(m, in) == 1 && GetBit(m, i) == 0) {
          ProbaMask[m] *= Proba[i][in];
          ProbaMask[m] %= MOD;
        }
        if (GetBit(m, in) == 0 && GetBit(m, i) == 1) {
          ProbaMask[m] *= Proba[in][i];
          ProbaMask[m] %= MOD;
        }
      }
    }
    long long ans = 0;
    for (long long i = 0; i < nbTeams; i++) {
      fill_n(DP, 1 << nbTeams, 0);
      for (long long mask = 1 << i; mask < (1 << nbTeams);
           mask = (mask + 1) | (1 << i)) {
        long long newMask = Transpo[mask];
        DP[mask] = ProbaMask[newMask];
        for (long long m = (mask - 1) & mask; m > 0; m = (m - 1) & mask) {
          if (!((m >> i) & 1)) continue;
          DP[mask] -= DP[m] * ProbaMask[newMask + Transpo[m]];
          DP[mask] %= MOD;
        }
        if (DP[mask] < 0) DP[mask] += MOD;
      }
      ans += DP[(1 << nbTeams) - 1];
    }
    printf("%lld\n", ans % MOD);
    return;
  }
  signed main() {
    long long nbTests = 1;
    while (nbTests--) {
      Solve();
    }
    return 0;
  }
  