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 pow_mod(int a, int k, int mod) {
    if (k == 0) {
      return 1;
    }
    int sub = pow_mod(a, k / 2, mod);
    if (k % 2 == 0) {
      return ((long long)sub * sub) % mod;
    } else {
      return (((long long)sub * sub) % mod * a) % mod;
    }
  }
  int inv_mod(int a, int p) { return pow_mod(a, p - 2, p); }
  int main() {
    int h, i, j, k;
    int t, n, m;
    long long mod = 1000000007;
    long long a[14];
    long long mem[1 << 14];
    long long part[1 << 14][14];
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    int b, c, bs;
    long long answer = 0;
    for (c = 1; c <= n; c++) {
      for (b = 1; b < (1 << n); b++) {
        int ones = 0;
        for (i = 0; i < n; i++) {
          if (b & (1 << i)) {
            ones++;
          }
        }
        if (ones == c) {
          mem[b] = 1;
          for (bs = 1; bs < (1 << n); bs++) {
            int check = 1;
            for (i = 0; i < n; i++) {
              if (!(b & (1 << i)) && bs & (1 << i)) {
                check = 0;
                break;
              }
            }
            if (check && b != bs) {
              long long prob = mem[bs];
              for (i = 0; i < n; i++) {
                if (b & (1 << i) && !(bs & (1 << i))) {
                  prob = (prob * part[bs][i]) % mod;
                }
              }
              mem[b] = (mem[b] - prob + mod) % mod;
            }
          }
          for (i = 0; i < n; i++) {
            part[b][i] = 1;
            if (!(b & (1 << i))) {
              for (j = 0; j < n; j++) {
                if (b & (1 << j)) {
                  part[b][i] = (part[b][i] * a[j]) % mod;
                  part[b][i] = (part[b][i] * inv_mod(a[i] + a[j], mod)) % mod;
                }
              }
            }
          }
          long long total = 1;
          for (i = 0; i < n; i++) {
            if (!(b & (1 << i))) {
              total = (total * part[b][i]) % mod;
            }
          }
          answer = (answer + (mem[b] * total) % mod * c) % mod;
        }
      }
    }
    printf("%lld", answer);
  }
  