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;
  template <typename T>
  inline void read(T &FF) {
    int RR = 1;
    FF = 0;
    char CH = getchar();
    for (; !isdigit(CH); CH = getchar())
      if (CH == '-') RR = -RR;
    for (; isdigit(CH); CH = getchar()) FF = FF * 10 + CH - 48;
    FF *= RR;
  }
  inline void file(string str) {
    freopen((str + ".in").c_str(), "r", stdin);
    freopen((str + ".out").c_str(), "w", stdout);
  }
  const long long mod = 1e9 + 7;
  const int N = 15, M = 5e4 + 10;
  int n, ai[N];
  long long g[M][N], f[M], flag[M];
  long long Qpow(long long x, long long k) {
    if (k <= 1) return k ? x : 1;
    long long h = Qpow(x, k / 2);
    h = h * h % mod;
    return (k & 1) ? h * x % mod : h;
  }
  long long suanf(int i) {
    long long p = 0;
    if (flag[i]) return f[i];
    flag[i] = 1;
    for (int j = (i - 1) & i; j; j = (j - 1) & i) {
      if (!j) break;
      long long c = 1;
      for (int k = 1; k <= n; k++)
        if ((j >> (k - 1)) & 1) (c *= g[i ^ j][k]) %= mod;
      p += c * suanf(j) % mod;
      p %= mod;
    }
    f[i] = (1 - p + mod) % mod;
    return f[i];
  }
  int main() {
    read(n);
    for (int i = 1; i <= n; i++) read(ai[i]);
    for (int i = 0; i <= (1 << n) - 1; i++)
      for (int j = 1; j <= n; j++) {
        g[i][j] = 1;
        for (int k = 1; k <= n; k++)
          if ((i >> (k - 1)) & 1)
            (g[i][j] *= ai[j] * Qpow(ai[j] + ai[k], mod - 2) % mod) %= mod;
      }
    long long res = 0;
    for (int i = 1; i <= (1 << n) - 1; i++) {
      long long c = 1, t = 0;
      for (int j = 1; j <= n; j++)
        if ((i >> (j - 1)) & 1) (c *= g[((1 << n) - 1) ^ i][j]) %= mod, t++;
      res += suanf(i) * c % mod * t % mod;
      res %= mod;
    }
    cout << res << endl;
    return 0;
  }
  