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;
  int recur_depth = 0;
  const long long sz = 14, isz = 2e6 + 10, mod = 1e9 + 7;
  long long p[sz], win[sz][sz], scc[(1 << sz)], n;
  long long fastPow(long long x, long long n, long long MOD) {
    long long ret = 1;
    while (n) {
      if (n & 1) ret = (ret * x) % MOD;
      x = (x * x) % MOD;
      n >>= 1;
    }
    return ret % MOD;
  }
  inline long long inv(long long x) { return fastPow(x, mod - 2, mod); }
  long long prob(long long from, long long to) {
    long long ret = 1;
    for (long long i = 0; i < n; ++i) {
      for (long long j = 0; j < n; ++j) {
        if ((from >> i & 1) && (to >> j & 1)) ret = (ret * win[i][j]) % mod;
      }
    }
    return ret;
  }
  int main() {
    std::ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    cin >> n;
    for (long long i = 0; i < n; ++i) cin >> p[i];
    for (long long i = 0; i < n; ++i)
      for (long long j = 0; j < n; ++j)
        win[i][j] = (p[i] * inv(p[i] + p[j])) % mod;
    long long ans = 0, lim = (1 << n) - 1;
    for (int mask = 0; mask <= lim; mask++) {
      scc[mask] = 1;
      for (int i = mask; i > 0; i = (i - 1) & mask) {
        if (i == mask) continue;
        scc[mask] -= (scc[i] * prob(i, mask ^ i)) % mod;
        if (scc[mask] < 0) scc[mask] += mod;
      }
      long long win_prob = (scc[mask] * __builtin_popcount(mask)) % mod;
      win_prob = (win_prob * prob(mask, lim ^ mask)) % mod;
      ans += win_prob;
      if (ans >= mod) ans -= mod;
    }
    cout << ans << '\n';
    return 0;
  }
  