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 n, ans;
  int a[19];
  int inv[2000009];
  int f[17009];
  int tag[19];
  int qpow(int a, int b) {
    int ans = 1;
    for (; b; b >>= 1, a = (long long)a * a % 1000000007)
      if (b & 1) ans = (long long)ans * a % 1000000007;
    return ans;
  }
  void init() {
    for (int i = 1; i <= 2000000; i++) inv[i] = qpow(i, 1000000007 - 2);
  }
  int main() {
    init();
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", a + i);
    for (int s = 1; s < (1 << n); s++) {
      int sm = 0;
      for (int t = (s - 1) & s; t; t = (t - 1) & s) {
        int p = s ^ t, pb = 1;
        for (int i = 1; i <= n; i++)
          if (t >> (i - 1) & 1)
            for (int j = 1; j <= n; j++)
              if (p >> (j - 1) & 1)
                pb = (long long)pb * a[i] % 1000000007 * inv[a[i] + a[j]] %
                     1000000007;
        sm = (sm + (long long)f[t] * pb) % 1000000007;
      }
      f[s] = (1000000007 + 1 - sm) % 1000000007;
    }
    int ans = 0;
    for (int i = 1; i <= n; i++)
      for (int s = 1; s < (1 << n); s++)
        if (s >> (i - 1) & 1) {
          int pb = 1;
          for (int i = 1; i <= n; i++)
            if (s >> (i - 1) & 1)
              for (int j = 1; j <= n; j++)
                if (!(s >> (j - 1) & 1))
                  pb = (long long)pb * a[i] % 1000000007 * inv[a[i] + a[j]] %
                       1000000007;
          ans = (ans + (long long)f[s] * pb) % 1000000007;
        }
    printf("%d", ans);
    return 0;
  }
  