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 maxS = (1 << 14), maxn = 18, maxa = 1e6 + 5;
  const long long MOD = 1e9 + 7;
  int n, mxS, a[maxn];
  bool inS(int S, int x) { return (S >> (x - 1)) & 1; }
  int chgS(int S, int x) { return S ^ (1 << (x - 1)); }
  long long p[maxn][maxS], f[maxn][maxS], ans, inv[maxa << 1];
  void init() {
    inv[1] = 1;
    for (int i = 2; i < (maxa << 1); i++)
      inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
    for (int i = 1; i <= n; i++)
      for (int S = 1; S <= mxS; S++)
        if (!inS(S, i)) {
          p[i][S] = 1;
          for (int j = 1; j <= n; j++)
            if (inS(S, j))
              p[i][S] = p[i][S] * a[j] % MOD * inv[a[i] + a[j]] % MOD;
        }
  }
  long long g(int S1, int S2) {
    long long ret = 1;
    for (int i = 1; i <= n; i++)
      if (inS(S1, i)) ret = ret * p[i][S2] % MOD;
    return ret;
  }
  int main() {
    scanf("%d", &n), mxS = (1 << n) - 1;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    init();
    for (int i = 1; i <= n; i++) {
      for (int S = 1; S <= mxS; S++)
        if (inS(S, i)) {
          f[i][S] = 1;
          for (int S1 = (S - 1) & S; S1; S1 = (S1 - 1) & S)
            if (inS(S1, i))
              f[i][S] = (f[i][S] - f[i][S1] * g(S1, S ^ S1) % MOD + MOD) % MOD;
        }
      ans = (ans + f[i][mxS]) % MOD;
    }
    printf("%lld\n", ans);
    return 0;
  }
  