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;
  using ll = long long;
  const ll MOD = (ll)1e9 + 7;
  ll mAdd(ll a, ll b) {
    if (a + b >= MOD)
      return a + b - MOD;
    else
      return a + b;
  }
  ll mSub(ll a, ll b) {
    if (a < b)
      return a + MOD - b;
    else
      return a - b;
  }
  ll modPow(ll a, ll b) {
    if (b & 1) return (a * modPow(a, b ^ 1)) % MOD;
    if (b == 0) return 1;
    return modPow(a * a % MOD, b >> 1);
  }
  const int N = 14;
  ll winp[N][N];
  ll mwinp[N][1 << N];
  ll sconp[1 << N];
  ll winAll(ll mask, ll sub) {
    ll res = 1;
    for (int i = 0; (1 << i) <= sub; ++i) {
      if (sub & (1 << i)) res = (res * mwinp[i][mask ^ sub]) % MOD;
    }
    return res;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    vector<ll> as(n);
    for (ll& a : as) cin >> a;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        winp[i][j] = (as[i] * modPow(mAdd(as[i], as[j]), MOD - 2)) % MOD;
      }
      for (int mask = 0; mask < (1 << n); ++mask) {
        ll res = 1;
        for (int j = 0; j < n; ++j) {
          if (mask & (1 << j)) res = (res * winp[i][j]) % MOD;
        }
        mwinp[i][mask] = res;
      }
    }
    ll ans = 0;
    ll full = (1 << n) - 1;
    for (int mask = 1; mask < (1 << n); ++mask) {
      sconp[mask] = 1;
      for (int sub = (mask - 1) & mask; sub; sub = (sub - 1) & mask) {
        ll prob = (sconp[sub] * winAll(mask, sub)) % MOD;
        sconp[mask] = mSub(sconp[mask], prob);
      }
      ll mult = (sconp[mask] * winAll(full, mask)) % MOD;
      ans = (ans + __builtin_popcount(mask) * mult) % MOD;
    }
    cout << ans << '\n';
  }
  