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;
  using ll = long long;
  const ll N = 17;
  const ll mod = 1e9 + 7;
  ll ksm(ll a, ll b) {
    ll s = 1;
    while (b) {
      if (b & 1) s = s * a % mod;
      a = a * a % mod;
      b >>= 1;
    }
    return s;
  }
  ll n, mx, a[N], p[N][1 << N], mp[N][N], dp[1 << N];
  inline ll calc(ll S, ll T) {
    ll ret = 1;
    for (int i = 0; i < n; i++) {
      if (S & 1 << i) ret = ret * p[i][T] % mod;
    }
    return ret;
  }
  int main() {
    cin >> n;
    mx = (1 << n) - 1;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (i == j) continue;
        mp[i][j] = a[i] * ksm(a[i] + a[j], mod - 2) % mod;
      }
    }
    for (int i = 0; i < n; i++) {
      for (int S = 0; S <= mx; S++) {
        p[i][S] = 1;
        if (S & 1 << i) continue;
        for (int j = 0; j < n; j++) {
          if (S & 1 << j) p[i][S] = p[i][S] * mp[i][j] % mod;
        }
      }
    }
    ll ans = 0;
    for (int S = 1; S <= mx; S++) {
      int cnt = 0, x = S;
      while (x) {
        cnt += x & 1;
        x >>= 1;
      }
      dp[S] = 1;
      for (int T = S & S - 1; T; T = S & T - 1) {
        dp[S] = (dp[S] - dp[T] * calc(T, S ^ T) % mod + mod) % mod;
      }
      ans = (ans + cnt * dp[S] % mod * calc(S, mx ^ S)) % mod;
    }
    cout << ans << endl;
    return 0;
  }
  