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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int rnd(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
  template <class T>
  void Min(T &a, const T b) {
    if (a > b) a = b;
  }
  template <class T>
  void Max(T &a, const T b) {
    if (a < b) a = b;
  }
  const int inf = 0x3f3f3f3f;
  const long long linf = 0x3f3f3f3f3f3f3f3f;
  const int N = 14;
  const long long mod = 1e9 + 7;
  int n;
  long long p[N][N], ar[N], f[1 << N];
  long long pm(long long a, long long b) {
    long long ret = 1;
    for (a %= mod; b; b >>= 1, a = a * a % mod)
      if (b & 1) ret = ret * a % mod;
    return ret;
  }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    ;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> ar[i];
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) {
        p[i][j] = ar[i] * pm(ar[i] + ar[j], mod - 2) % mod;
      }
    for (int s = 1; s < 1 << n; s++) {
      f[s] = 1;
      long long sub = 0;
      for (int t = s - 1 & s; t; t = t - 1 & s) {
        long long a = f[t];
        for (int i = 0; i < n; i++)
          if (t & 1 << i) {
            for (int j = 0; j < n; j++)
              if ((t & 1 << j) == 0 && s & 1 << j) {
                a = a * p[i][j] % mod;
              }
          }
        sub = (sub + a) % mod;
      }
      f[s] = (f[s] - sub + mod) % mod;
    }
    long long ans = 0;
    for (int s = 0; s < 1 << n; s++) {
      long long cnt = 0, add = f[s];
      for (int i = 0; i < n; i++)
        if (s & 1 << i) {
          cnt++;
          for (int j = 0; j < n; j++)
            if ((s & 1 << j) == 0) {
              add = add * p[i][j] % mod;
            }
        }
      add = add * cnt % mod;
      ans = (ans + add) % mod;
    }
    cout << ans;
  }
  