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 long long mod = 1000000007ll;
  const int N = 14;
  long long power(long long a, long long x) {
    long long ans = 1;
    while (x) {
      if (x & 1) ans = ans * a % mod;
      a = a * a % mod;
      x >>= 1;
    }
    return ans;
  }
  long long inv(long long x) { return power(x, mod - 2); }
  long long h[N + 5][1 << N];
  long long p[N + 5][N + 5];
  long long a[N + 5], f[1 << N];
  long long g[N + 5];
  int cal(int x) {
    int ans = 0;
    while (x) {
      ans += x & 1;
      x >>= 1;
    }
    return ans;
  }
  long long temp[1 << N];
  int main() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) p[i][j] = a[i] * inv(a[i] + a[j]) % mod;
    for (int i = 1; i <= n; i++)
      for (int j = 0; j < (1 << n); j++) {
        if (j & (1 << (i - 1))) continue;
        h[i][j] = 1;
        for (int k = 1; k <= n; k++)
          if (j & (1 << (k - 1))) h[i][j] = h[i][j] * p[i][k] % mod;
      }
    long long ans = 0;
    for (int i = 1; i < 1 << n; i++) {
      g[i] = 1;
      for (int j = i & (i - 1); j; j = (j - 1) & i) {
        for (int k = 1; k <= n; k++)
          if ((i & (1 << (k - 1))) && !(j & (1 << (k - 1)))) {
            g[j] = g[j ^ (1 << (k - 1))] * h[k][((1 << n) - 1) ^ i] % mod;
            break;
          }
        f[i] = (f[i] + (mod - temp[j]) * g[j] % mod) % mod;
      }
      for (int k = 1; k <= n; k++)
        if ((i & (1 << (k - 1)))) {
          g[0] = g[(1 << (k - 1))] * h[k][((1 << n) - 1) ^ i] % mod;
          break;
        }
      f[i] = (f[i] + g[0]) % mod;
      temp[i] = f[i];
      ans = (ans + f[i] * cal(i) % mod) % mod;
    }
    printf("%lld", ans);
    return 0;
  }
  