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 const N = 14;
  int const mod = 1e9 + 7;
  long long ext_gcd(long long a, long long b, long long &x, long long &y) {
    long long t, ret;
    if (!b) {
      x = 1, y = 0;
      return a;
    }
    ret = ext_gcd(b, a % b, x, y);
    t = x, x = y, y = t - a / b * y;
    return ret;
  }
  long long inv(long long a, long long b, long long c) {
    long long x, y;
    ext_gcd(a, c, x, y);
    return (x * b % c + c) % c;
  }
  int n;
  long long a[N];
  long long ff[1 << N], ga[1 << N][N + 1], gb[1 << N];
  long long g(int x, int y) {
    long long da = ga[x][__builtin_popcount(y)];
    long long db = inv(gb[x ^ y], gb[x] * gb[y] % mod, mod);
    return da * db % mod;
  }
  long long f(int msk, int all) {
    if (ff[msk] != -1) return ff[msk];
    long long ans = g(msk, all ^ msk);
    for (int i = (msk - 1) & msk; i > 0; i = (i - 1) & msk) {
      ans -= f(i, all) * g(msk ^ i, all ^ msk);
      ans %= mod;
    }
    return ff[msk] = ans;
  }
  void ginit(int n) {
    for (int i = 0; i < (1 << n); ++i) {
      ga[i][0] = gb[i] = 1;
      long long z = 1;
      for (int j = 0; j < (n); ++j)
        if (i >> j & 1) z = z * a[j] % mod;
      for (int j = 1; j <= (n); ++j) ga[i][j] = ga[i][j - 1] * z % mod;
      for (int j = 0; j < (n); ++j)
        for (int k = j + 1; k < n; ++k) {
          if ((i >> j & 1) && (i >> k & 1)) {
            gb[i] = gb[i] * (a[j] + a[k]) % mod;
          }
        }
    }
  }
  int main() {
    scanf("%d", &n);
    for (int i = 0; i < (n); ++i) scanf("%lld", &a[i]);
    for (int i = 0; i < (1 << n); ++i) ff[i] = -1;
    ginit(n);
    long long ans = 0;
    int all = (1 << n) - 1;
    for (int i = 1; i <= (all); ++i) {
      ans += f(i, all) * __builtin_popcount(i);
      ans %= mod;
    }
    ans = (ans + mod) % mod;
    printf("%lld\n", ans);
    return 0;
  }
  