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;
  void abc() { cout << endl; }
  template <typename T, typename... U>
  void abc(T a, U... b) {
    cout << a << ' ', abc(b...);
  }
  template <typename T>
  void printv(T l, T r) {
    while (l != r) cout << *l << " \n"[++l == r];
  }
  template <typename A, typename B>
  istream& operator>>(istream& o, pair<A, B>& a) {
    return o >> a.first >> a.second;
  }
  template <typename A, typename B>
  ostream& operator<<(ostream& o, pair<A, B> a) {
    return o << '(' << a.first << ", " << a.second << ')';
  }
  template <typename T>
  ostream& operator<<(ostream& o, vector<T> a) {
    bool is = false;
    for (T i : a) {
      o << (is ? ' ' : '{'), is = true, o << i;
    }
    return o << '}';
  }
  template <typename T>
  struct vv : vector<vector<T>> {
    vv(int n, int m, T v) : vector<vector<T>>(n, vector<T>(m, v)) {}
    vv() {}
  };
  template <typename T>
  struct vvv : vector<vv<T>> {
    vvv(int n, int m, int k, T v) : vector<vv<T>>(n, vv<T>(m, k, v)) {}
    vvv() {}
  };
  const int mod = 1e9 + 7, N = 200001, logN = 20, K = 80000;
  long long int modpow(long long int a, int b) {
    long long int ans = 1;
    for (; b; b >>= 1, a = a * a % mod) {
      if (b & 1) ans = ans * a % mod;
    }
    return ans;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    vector<int> a(n);
    vv<int> inv(n, n, 0);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
    }
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j) {
        inv[i][j] = modpow((a[i] + a[j]) % mod, mod - 2);
      }
    vector<long long int> dp(1 << n, 1);
    for (int s = 0; s < 1 << n; ++s) {
      vector<int> v;
      for (int i = 0; i < n; ++i)
        if (s >> i & 1) v.push_back(i);
      for (int t = s ^ (s & (-s)); t > 0; t = (t - 1) & s) {
        long long int res = 1;
        for (int i : v)
          if (t >> i & 1) {
            for (int j : v)
              if (~t >> j & 1) {
                res = res * a[j] % mod * inv[i][j] % mod;
              }
          }
        dp[s] -= dp[t] * res % mod;
        if (dp[s] < 0) dp[s] += mod;
      }
    }
    long long int ans = 0;
    for (int s = 0; s < 1 << n; ++s) {
      int k = __builtin_popcount(s);
      long long int res = dp[s];
      for (int i = 0; i < n; ++i)
        if (s >> i & 1) {
          for (int j = 0; j < n; ++j)
            if (~s >> j & 1) {
              res = res * a[i] % mod * inv[i][j] % mod;
            }
        }
      ans += res * k % mod;
      if (ans >= mod) ans -= mod;
    }
    cout << ans << endl;
  }
  