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;
  const long long INFLL = 1e18;
  const int INF = 1e9 + 1;
  const int mod = 1e9 + 7;
  long long mul(long long a, long long b) { return (a * b) % mod; }
  long long add(long long a, long long b) { return ((a + b) % mod + mod) % mod; }
  long long my_pow(long long n, long long m) {
    if (m == 0) return 1;
    long long now = my_pow(n, m / 2);
    if (m % 2 == 0)
      return (now * now) % mod;
    else
      return (((now * now) % mod) * n) % mod;
  }
  long long rev(long long a) { return my_pow(a, mod - 2); }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    vector<long long> a(n);
    for (auto &c : a) cin >> c;
    vector<vector<long long>> h(n, vector<long long>(1 << n));
    for (int i = 0; i < n; i++) {
      for (int mask = 0; mask < (1 << n); mask++) {
        if (mask & (1 << i)) continue;
        h[i][mask] = 1;
        for (int j = 0; j < n; j++) {
          if (mask & (1 << j))
            h[i][mask] = mul(h[i][mask], mul(a[i], rev(add(a[i], a[j]))));
        }
      }
    }
    vector<long long> f(1 << n);
    long long ans = 0;
    for (int mask = 1; mask < (1 << n); mask++) {
      long long x = 1;
      for (int j = 0; j < n; j++) {
        if (!(mask & (1 << j))) continue;
        x = mul(x, h[j][((1 << n) - 1) ^ mask]);
      }
      long long y = 1;
      for (int mask1 = mask; mask1 > 0; mask1 = (mask1 - 1) & mask) {
        if (mask1 == mask) continue;
        long long z = 1;
        for (int j = 0; j < n; j++) {
          if (!(mask1 & (1 << j))) continue;
          z = mul(z, h[j][((1 << n) - 1) ^ mask]);
        }
        y = add(y, -mul(f[mask1], rev(z)));
      }
      f[mask] = mul(x, y);
      ans = add(ans, mul(f[mask], __builtin_popcount(mask)));
    }
    cout << ans;
    return 0;
  }
  