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;
  long long power[15];
  long long prob[15][15];
  long long py[1 << 15][15];
  long long dp[1 << 15];
  const long long mod = 1e9 + 7;
  long long ksm(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b & 1) ans = ans * a % mod;
      a = a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> power[i];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        prob[i][j] = power[i] * ksm(power[i] + power[j], mod - 2) % mod;
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < 1 << n; j++) {
        long long cur = 1;
        for (int k = 0; k < n; k++) {
          if (j & (1 << k)) cur = cur * prob[k][i] % mod;
        }
        py[j][i] = cur;
      }
    }
    for (int i = 1; i < 1 << n; i++) {
      long long res = 1;
      vector<int> pos;
      for (int j = 0; j < n; j++) {
        if (i & (1 << j)) pos.push_back(j);
      }
      int sz = pos.size();
      for (int j = 1; j < (1 << sz) - 1; j++) {
        int val = 0;
        for (int k = 0; k < sz; k++) {
          if (j & (1 << k)) val += 1 << pos[k];
        }
        long long cur = dp[val];
        for (int k = 0; k < sz; k++) {
          if (!(j & (1 << k))) cur = cur * py[val][pos[k]] % mod;
        }
        res = (res - cur) % mod;
      }
      dp[i] = res;
    }
    long long ans = 0;
    for (int i = 0; i < 1 << n; i++) {
      long long cur = 1;
      for (int j = 0; j < n; j++) {
        if (!(i & (1 << j))) cur = cur * py[i][j] % mod;
      }
      ans = (ans + (long long)__builtin_popcount(i) * dp[i] % mod * cur) % mod;
    }
    cout << (ans + mod) % mod << '\n';
  }
  