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>
  #pragma GCC optimize("O3")
  using namespace std;
  void setIO() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
  }
  const long long N = 2e1 + 0;
  const long long M = 5e2 + 5;
  const long long inf = 9e17;
  const long long mod = 1e9 + 7;
  long long xyz = 1;
  long long n;
  long long arr[N];
  long long pre[N][1 << N];
  long long dp[1 << N];
  long long sqr(long long x) { return x * x % mod; }
  long long bin(long long b, long long e) {
    if (e == 0) return 1;
    if (e == 1) return b;
    return sqr(bin(b, e / 2)) * bin(b, e % 2) % mod;
  }
  long long inv(long long x) { return bin(x % mod, mod - 2); }
  void run() {
    cin >> n;
    for (long long i = 0; i < n; i++) cin >> arr[i];
    for (long long m = 0; m < (1 << n); m++) {
      for (long long i = 0; i < n; i++) {
        if (m >> i & 1) continue;
        pre[i][m] = 1;
        for (long long j = 0; j < n; j++) {
          if (m >> j & 1) {
            pre[i][m] *= arr[i] * inv(arr[i] + arr[j]) % mod;
            pre[i][m] %= mod;
          }
        }
      }
    }
    long long ans = 0;
    for (long long m = 1; m < (1 << n); m++) {
      long long cnt = __builtin_popcount(m);
      long long cur = 1;
      for (long long i = 0; i < n; i++) {
        if (m >> i & 1) {
          cur *= pre[i][((1 << n) - 1) & (~m)];
          cur %= mod;
        }
      }
      dp[m] = 1;
      for (long long s = m; s; s = (s - 1) & m) {
        if (s == m) continue;
        long long can = dp[s];
        for (long long i = 0; i < n; i++) {
          if (s >> i & 1) {
            can *= pre[i][m ^ s] % mod;
            can %= mod;
          }
        }
        dp[m] += mod - can;
        dp[m] %= mod;
      }
      ans += cur * dp[m] % mod * cnt % mod;
      ans %= mod;
    }
    cout << ans << "\n";
  }
  signed main() {
    setIO();
    while (xyz--) run();
    return 0;
  }
  