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 inv(int v) {
    return v <= 1 ? v
                  : (1000000007) - (long long)inv((1000000007) % v) *
                                       ((1000000007) / v) % (1000000007);
  }
  struct mint {
    int v;
    mint() = default;
    mint(long long v) : v((v % (1000000007) + (1000000007)) % (1000000007)) {}
    mint& operator+=(mint a) {
      v = (v + a.v) % (1000000007);
      return *this;
    }
    mint& operator-=(mint a) {
      v = (v - a.v + (1000000007)) % (1000000007);
      return *this;
    }
    mint& operator*=(mint a) {
      v = ((long long)v * a.v) % (1000000007);
      return *this;
    }
    mint& operator/=(mint a) {
      v = ((long long)v * inv(a.v)) % (1000000007);
      return *this;
    }
    friend mint operator+(mint a, mint b) { return mint(a.v + b.v); }
    friend mint operator-(mint a, mint b) { return mint(a.v - b.v); }
    friend mint operator*(mint a, mint b) { return mint((long long)a.v * b.v); }
    friend mint operator/(mint a, mint b) {
      return mint((long long)a.v * inv(b.v));
    }
  };
  mint pow(mint base, long long exponent) {
    if (exponent <= 0) {
      return 1;
    }
    mint ret = 1;
    while (exponent) {
      if (exponent & 1) {
        ret *= base;
      }
      exponent >>= 1;
      base = base * base;
    }
    return ret;
  }
  int main() {
    std::ios::sync_with_stdio(false);
    int n;
    while (cin >> n) {
      vector<int> w(n);
      for (int i = 0; i < n; ++i) {
        cin >> w[i];
      }
      vector<vector<mint>> a(n, vector<mint>(n));
      for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
          a[i][j] = mint(w[i]) / mint(w[i] + w[j]);
        }
      }
      vector<vector<mint>> g(1 << n, vector<mint>(n));
      for (int mask = 0; mask < (1 << n); ++mask) {
        for (int i = 0; i < n; ++i) {
          g[mask][i] = 1;
          for (int j = 0; j < n; ++j) {
            if (mask & (1 << j)) {
              g[mask][i] *= a[i][j];
            }
          }
        }
      }
      vector<mint> f(1 << n);
      for (int mask = 1; mask < (1 << n); ++mask) {
        f[mask] = 1;
        for (int sub_mask = mask; sub_mask > 0;
             sub_mask = ((sub_mask - 1) & mask)) {
          if (sub_mask < mask) {
            mint val = f[sub_mask];
            int other_mask = mask - sub_mask;
            for (int i = 0; i < n; ++i) {
              if (sub_mask & (1 << i)) {
                val *= g[other_mask][i];
              }
            }
            f[mask] -= val;
          }
        }
      }
      mint result = 0;
      for (int mask = 1; mask < (1 << n); ++mask) {
        int count = 0;
        for (int i = 0; i < n; ++i) {
          if (mask & (1 << i)) {
            ++count;
          }
        }
        mint val = f[mask] * mint(count);
        int other_mask = (1 << n) - 1 - mask;
        for (int i = 0; i < n; ++i) {
          if (mask & (1 << i)) {
            val *= g[other_mask][i];
          }
        }
        result += val;
      }
      printf("%d\n", result.v);
    }
    return 0;
  }
  