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;
  long long read() {
    long long x = 0, f = 1;
    char ch = ' ';
    while (!isdigit(ch)) {
      ch = getchar();
      if (ch == '-') f = -1;
    }
    while (isdigit(ch)) x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x * f;
  }
  const int mod = 1e9 + 7;
  int n, a[15], f[1 << 14], g[1 << 14];
  int ni[15][15];
  inline int P(int s, int t) {
    int res = 1;
    for (int i = 0; i < n; i++) {
      if (!(s & (1 << i))) continue;
      for (int j = 0; j < n; j++) {
        if (!(t & (1 << j))) continue;
        res = 1ll * res * ni[i][j] % mod;
      }
    }
    return res;
  }
  int fpow(int n) {
    int res = 1, m = mod - 2;
    while (m) {
      if (m & 1) res = 1ll * res * n % mod;
      n = 1ll * n * n % mod;
      m >>= 1;
    }
    return res;
  }
  int main() {
    cin >> n;
    int S = (1 << n) - 1;
    for (int i = 0; i < n; i++) a[i] = read();
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        ni[i][j] = 1ll * a[i] * fpow(a[i] + a[j]) % mod;
      }
    }
    for (int i = 0; i <= S; i++) {
      long long s = S - i;
      g[i] = P(i, s);
    }
    int ans = 0;
    for (int i = 0; i <= S; i++) {
      f[i] = g[i];
      for (int j = (i - 1) & i; j; j = (j - 1) & i) {
        f[i] -= 1ll * f[j] * g[i - j] % mod * fpow(P(i - j, j)) % mod;
        if (f[i] < 0) f[i] += mod;
      }
      ans += 1ll * f[i] * __builtin_popcount(i) % mod;
      if (ans > mod) ans -= mod;
    }
    cout << ans;
  }
  