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>
  int read() {
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') x = x * 10 + (c - '0'), c = getchar();
    return f * x;
  }
  long long min(long long a, long long b) { return a < b ? a : b; }
  long long max(long long a, long long b) { return a > b ? a : b; }
  template <typename T>
  bool umin(T& a, T t) {
    if (a > t) return a = t, 1;
    return 0;
  }
  template <typename T>
  bool umax(T& a, T t) {
    if (a < t) return a = t, 1;
    return 0;
  }
  int abs(int x) { return x > 0 ? x : -x; }
  const int MAXN = 14, mod = 1e9 + 7;
  int f[1 << MAXN], a[MAXN];
  int res[1 << MAXN];
  int inv[2000011];
  void init() {
    inv[1] = 1;
    for (int i = 2; i <= 2000000; ++i)
      inv[i] = long long(mod - mod / i) * inv[mod % i] % mod;
  }
  int main() {
    init();
    int n = read(), all = (1 << n) - 1;
    for (int i = 0; i < n; ++i) a[i] = read();
    for (int s = 1; s < (1 << n); ++s) {
      f[s] = 1;
      for (int t = (s - 1) & s; t; t = (t - 1) & s) {
        int p = f[t];
        for (int i = 0; i < n; ++i)
          if (t & (1 << i))
            for (int j = 0; j < n; ++j)
              if ((s & (1 << j)) && !(t & (1 << j)))
                p = long long(p) * a[i] % mod * inv[a[i] + a[j]] % mod;
        f[s] = (f[s] - p) % mod;
      }
    }
    int ans = 0;
    for (int s = 1; s < (1 << n); ++s) {
      int cnt = __builtin_popcount(s), p = f[s];
      for (int i = 0; i < n; ++i)
        if (s & (1 << i))
          for (int j = 0; j < n; ++j)
            if (!(s & (1 << j)))
              p = long long(p) * a[i] % mod * inv[a[i] + a[j]] % mod;
      ans = (ans + long long(p) * cnt) % mod;
    }
    printf("%d\n", (ans + mod) % mod);
    return 0;
  }
  