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;
  int i, j, m, n, u, v;
  long long a[15], mod, lose[14][16389], f[16389], g[16389][15];
  long long sz[16389], inv[16389];
  long long getinv(long long a, long long p) {
    if (a == 1) return 1;
    long long k = a - getinv(p % a, a);
    return (k * p + 1) / a;
  }
  long long group_lose(int v, int u) {
    int i, j;
    long long ret;
    ret = f[v] * f[u - v] % mod * g[u - v][sz[v]] % mod * inv[u] % mod;
    return ret;
  }
  int main() {
    mod = 1000000007;
    scanf("%d", &n);
    for (i = 0; i <= n - 1; i++) {
      scanf("%lld", &a[i]);
    }
    for (u = 0; u <= (1 << n) - 1; u++) {
      sz[u] = 0;
      for (i = 0; i <= n - 1; i++)
        if (u & (1 << i)) sz[u]++;
    }
    for (u = 0; u <= (1 << n) - 1; u++) {
      f[u] = 1;
      for (i = 0; i <= n - 1; i++)
        if (u & (1 << i)) {
          for (j = i + 1; j <= n - 1; j++)
            if (u & (1 << j)) {
              f[u] = f[u] * (a[i] + a[j]) % mod;
            }
        }
      inv[u] = getinv(f[u], mod);
    }
    for (u = 0; u <= (1 << n) - 1; u++) {
      g[u][0] = 1;
      g[u][1] = 1;
      for (i = 0; i <= n - 1; i++)
        if (u & (1 << i)) {
          g[u][1] = g[u][1] * a[i] % mod;
        }
      for (i = 2; i <= n; i++) {
        g[u][i] = g[u][i - 1] * g[u][1] % mod;
      }
    }
    for (u = 0; u <= (1 << n) - 1; u++) {
      for (i = 0; i <= n - 1; i++) {
        if (u & (1 << i)) {
          lose[i][u] = 0;
          for (v = u; v; v = (v - 1) & u)
            if ((v != u) && (v & (1 << i))) {
              long long temp = group_lose(v, u);
              lose[i][u] += (1 - lose[i][v] + mod) * temp % mod;
            }
          lose[i][u] %= mod;
        }
      }
    }
    long long ans = 0;
    for (i = 0; i <= n - 1; i++) {
      ans += (1 - lose[i][(1 << n) - 1] + mod);
    }
    ans %= mod;
    printf("%lld\n", ans);
    return 0;
  }
  