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 int a[16];
  long long int dp[1000000];
  long long int cost[16][16];
  long long int modpow(long long int base, int exp) {
    if (exp == 0) return 1;
    if (exp == 1) return base;
    long long int pv = modpow(base, exp / 2);
    pv = (pv * pv) % int(1e9 + 7);
    if (exp % 2 == 0) return pv;
    pv = (base * pv) % int(1e9 + 7);
    return pv;
  }
  int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        cost[i][j] =
            (a[i] * modpow(a[i] + a[j], int(1e9 + 7) - 2)) % int(1e9 + 7);
      }
    }
    for (int msk = 1; msk < (1 << n); msk++) {
      if (__builtin_popcount(msk) == 1) {
        dp[msk] = 1;
        continue;
      }
      vector<int> v;
      v.clear();
      for (int i = 0; i < n; i++) {
        int x = msk & (1 << i);
        if (x > 0) v.push_back(i);
      }
      int sz = v.size();
      dp[msk] = 1;
      for (int submask = 1; submask < (1 << sz) - 1; submask++) {
        long long int naive = 1;
        int prev_mask = 0;
        vector<int> inside_bits, outside_bits;
        inside_bits.clear();
        outside_bits.clear();
        for (int i = 0; i < sz; i++) {
          int x = submask & (1 << i);
          if (x > 0) {
            inside_bits.push_back(v[i]);
            prev_mask += (1 << v[i]);
          } else
            outside_bits.push_back(v[i]);
        }
        assert(!inside_bits.empty());
        assert(!outside_bits.empty());
        for (int in : inside_bits) {
          for (int out : outside_bits) {
            naive = naive * cost[in][out] % int(1e9 + 7);
          }
        }
        assert(prev_mask < (1 << n));
        dp[msk] -= (dp[prev_mask] * naive) % int(1e9 + 7);
        dp[msk] %= int(1e9 + 7);
      }
    }
    long long int ret = 0;
    for (int msk = 1; msk < (1 << n); msk++) {
      long long int naive = 1;
      for (int t = 0; t < n; t++) {
        for (int q = 0; q < n; q++) {
          int x = msk & (1 << t);
          int y = msk & (1 << q);
          if (x > 0 and y == 0) naive = (naive * cost[t][q]) % int(1e9 + 7);
        }
      }
      ret = (ret + (((__builtin_popcount(msk) * naive) % int(1e9 + 7)) * dp[msk] %
                    int(1e9 + 7))) %
            int(1e9 + 7);
    }
    if (ret < 0) ret += int(1e9 + 7);
    cout << ret << endl;
  }
  