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>
  inline int fmo(int x) { return x + ((x >> 31) & 1000000007); }
  inline int fp(int x, int k = 1000000007 - 2) {
    int res = 1;
    for (; k; k >>= 1, x = 1ll * x * x % 1000000007)
      if (k & 1) res = 1ll * res * x % 1000000007;
    return res;
  }
  int n, a[15], w[15][15];
  int p[1 << 15][15];
  std::unordered_map<long long, int> f;
  int g[1 << 15];
  int ans;
  int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        if (i != j) w[i][j] = 1ll * a[i] * fp(a[i] + a[j]) % 1000000007;
    for (int s = 0; s < (1 << n); s++)
      for (int i = 0; i < n; i++)
        if (!((s >> i) & 1)) {
          p[s][i] = 1;
          for (int j = 0; j < n; j++)
            if ((s >> j) & 1) p[s][i] = 1ll * p[s][i] * w[i][j] % 1000000007;
        }
    for (int s = 0; s < (1 << n); s++) {
      int x = ((1 << n) - 1) ^ s;
      for (int t = x;; t = (t - 1) & x) {
        int tmp = 1;
        for (int i = 0; i < n; i++)
          if ((s >> i) & 1) tmp = 1ll * tmp * p[t][i] % 1000000007;
        f[s << n | t] = tmp;
        if (!t) break;
      }
    }
    g[0] = 0;
    for (int s = 1; s < (1 << n); s++) {
      int tmp = 1;
      for (int t = (s - 1) & s;; t = (t - 1) & s) {
        tmp = fmo(tmp - 1ll * g[t] * f[t << n | (s ^ t)] % 1000000007);
        if (!t) break;
      }
      g[s] = tmp;
      ans = fmo(ans +
                1ll * __builtin_popcount(s) * g[s] % 1000000007 *
                    f[s << n | (((1 << n) - 1) ^ s)] % 1000000007 -
                1000000007);
    }
    printf("%d\n", ans);
  }
  