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;
  template <class T>
  using v2d = vector<vector<T> >;
  template <class T>
  bool uin(T &a, T b) {
    return a > b ? (a = b, true) : false;
  }
  template <class T>
  bool uax(T &a, T b) {
    return a < b ? (a = b, true) : false;
  }
  mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
  const int maxN = 14;
  const long long mod = 1e9 + 7;
  int n;
  long long a[14], f[1 << 14], w[14][14], h[14][1 << 14];
  long long power(long long a, long long b) {
    long long r = 1;
    while (b) {
      if (b & 1) {
        r = r * a % mod;
      }
      b /= 2;
      a = a * a % mod;
    }
    return r;
  }
  void solve() {
    cin >> n;
    for (int i = 0; i < (int)(n); ++i) {
      cin >> a[i];
    }
    for (int i = 0; i < (int)(n); ++i) {
      for (int j = 0; j < (int)(n); ++j) {
        w[i][j] = a[i] * power(a[i] + a[j], mod - 2) % mod;
      }
      for (int mask = 0; mask < (int)(1 << n); ++mask) {
        h[i][mask] = 1;
        for (int j = 0; j < (int)(n); ++j) {
          if (mask >> j & 1) {
            h[i][mask] = h[i][mask] * w[i][j] % mod;
          }
        }
      }
    }
    int m = (1 << n) - 1;
    for (int mask = 1; mask <= (int)(m); ++mask) {
      f[mask] = 1;
      int submask = (mask - 1) & mask;
      while (submask) {
        long long p = 1;
        for (int i = 0; i < (int)(n); ++i) {
          if (submask >> i & 1) {
            p = p * h[i][mask ^ submask] % mod;
          }
        }
        f[mask] = (f[mask] - p * f[submask] + mod * mod) % mod;
        submask = (submask - 1) & mask;
      }
    }
    long long ans = 0;
    for (int mask = 1; mask <= (int)(m); ++mask) {
      int cnt = __builtin_popcount(mask);
      long long p = 1;
      for (int i = 0; i < (int)(n); ++i) {
        if (mask >> i & 1) {
          p = p * h[i][m ^ mask] % mod;
        }
      }
      ans = (ans + cnt * f[mask] % mod * p) % mod;
    }
    cout << ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int T = 1;
    while (T--) {
      solve();
    }
    return 0;
  }
  