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;
  const int mod = 1e9 + 7;
  const int MAXN = 15;
  int n, a[MAXN], aa[MAXN][MAXN];
  int dp[1 << MAXN];
  int x[1 << MAXN][MAXN], f[1 << MAXN], g[1 << MAXN];
  int cnt[1 << MAXN];
  int power(int x, int n) {
    int res = 1;
    while (n) {
      if (n & 1) res = 1LL * res * x % mod;
      n >>= 1;
      x = 1LL * x * x % mod;
    }
    return res;
  }
  void solve() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    for (int i = 0; i < n; i++)
      for (int j = i + 1; j < n; j++) aa[i][j] = power(a[i] + a[j], mod - 2);
    f[0] = 1;
    for (int i = 1; i < 1 << n; i++) {
      cnt[i] = 0;
      x[i][0] = x[i][1] = f[i] = g[i] = 1;
      for (int j = 0; j < n; j++) {
        if (i & (1 << j)) {
          cnt[i]++;
          x[i][1] = 1LL * x[i][1] * a[j] % mod;
          for (int k = j + 1; k < n; k++)
            if (i & (1 << k)) {
              f[i] = 1LL * f[i] * (a[j] + a[k]) % mod;
              g[i] = 1LL * g[i] * aa[j][k] % mod;
            }
        }
      }
      for (int j = 2; j < n; j++) x[i][j] = 1LL * x[i][j - 1] * x[i][1] % mod;
    }
    int ans = 0, all = (1 << n) - 1;
    for (int i = 1; i < 1 << n; i++) {
      int p = 0;
      for (int j = (i - 1) & i; j; j = (j - 1) & i) {
        p = (p + 1LL * dp[j] * x[j][cnt[i] - cnt[j]] % mod * g[i] % mod * f[j] %
                     mod * f[i ^ j] % mod) %
            mod;
      }
      dp[i] = (1 - p + mod) % mod;
      ans = (ans + 1LL * cnt[i] * dp[i] % mod * x[i][cnt[all] - cnt[i]] % mod *
                       g[all] % mod * f[i] % mod * f[all ^ i] % mod) %
            mod;
    }
    printf("%d\n", ans);
  }
  int main() {
    int T = 1;
    for (int i = 1; i <= T; i++) solve();
    return 0;
  }
  