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 exp(long long x, long long y) {
    long long MOD = 1000000007;
    if (y == 1) {
      return x;
    }
    long long z = exp(x, y / 2);
    z = (z * z) % MOD;
    if (y % 2 == 1) {
      z = (z * x) % MOD;
    }
    return z;
  }
  long long inv(long long x) { return exp(x, 1000000005); }
  int main() {
    long long n;
    cin >> n;
    long long MOD = 1000000007;
    long long a[n];
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long b[n][n];
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < n; j++) {
        b[i][j] = (a[j] * inv(a[i] + a[j])) % MOD;
      }
    }
    long long m = 1 << n;
    long long c[n][m];
    for (long long i = 0; i < n; i++) {
      c[i][0] = 1;
      for (long long j = 0; j < n; j++) {
        for (long long k = 0; k < (1 << j); k++) {
          c[i][k + (1 << j)] = (c[i][k] * b[i][j]) % MOD;
        }
      }
    }
    long long d[m];
    d[0] = 0;
    for (long long i = 1; i < m; i++) {
      bool e[n];
      long long t = 0;
      for (long long j = 0; j < n; j++) {
        if ((i & (1 << j)) == 0) {
          e[j] = false;
        } else {
          e[j] = true;
          t++;
        }
      }
      long long g[t];
      long long temp = 0;
      for (long long j = 0; j < n; j++) {
        if (e[j]) {
          g[temp] = j;
          temp++;
        }
      }
      long long f[1 << t];
      f[0] = 0;
      for (long long j = 0; j < t; j++) {
        for (long long k = 0; k < (1 << j); k++) {
          f[k + (1 << j)] = f[k] + (1 << g[j]);
        }
      }
      long long v = 0;
      for (long long j = 0; j < (1 << t) - 1; j++) {
        long long w = d[f[j]];
        for (long long k = 0; k < n; k++) {
          if (e[k] && ((f[j] & (1 << k)) == 0)) {
            w = (w * c[k][f[j]]) % MOD;
          }
        }
        v = (v + w) % MOD;
      }
      d[i] = (MOD + 1 - v) % MOD;
    }
    long long ans = 0;
    for (long long i = 0; i < m; i++) {
      long long w = d[i];
      long long t = 0;
      for (long long k = 0; k < n; k++) {
        if ((i & (1 << k)) == 0) {
          w = (w * c[k][i]) % MOD;
        } else {
          t++;
        }
      }
      ans = (ans + (w * t)) % MOD;
    }
    cout << ans << '\n';
    return 0;
  }
  