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;
  using ld = long double;
  using pii = pair<int, int>;
  const int maxn = 1e5 + 10;
  const int mod = 1e9 + 7;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  ll pwr(ll a, ll pw) {
    ll ret = 1;
    while (pw > 0) {
      if (pw & 1) {
        ret = ret * a % mod;
      }
      a = a * a % mod;
      pw >>= 1;
    }
    return ret;
  }
  ll modinv(ll a) { return pwr(a, mod - 2); }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int n;
    cin >> n;
    vector<ll> v(n);
    for (int i = 0; i < n; i++) {
      cin >> v[i];
    }
    vector<vector<ll> > prob_win(n, vector<ll>(n));
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        prob_win[i][j] = v[i] * modinv(v[i] + v[j]) % mod;
        prob_win[j][i] = v[j] * modinv(v[i] + v[j]) % mod;
      }
    }
    vector<vector<ll> > win_msk(n, vector<ll>(1 << n, 1ll));
    for (int i = 0; i < n; i++) {
      for (int msk = 0; msk < 1 << n; msk++) {
        for (int bit = 0; bit < n; bit++) {
          if (msk >> bit & 1) {
            win_msk[i][msk] = win_msk[i][msk] * prob_win[i][bit] % mod;
          }
        }
      }
    }
    ll ans = 0;
    vector<ll> dp_prob(1 << n, 1ll);
    for (int msk = 1; msk < 1 << n; msk++) {
      ll num_winners = __builtin_popcount(msk);
      for (int bit = 0; bit < n; bit++) {
        if (msk >> bit & 1) {
          dp_prob[msk] = dp_prob[msk] * win_msk[bit][(1 << n) - 1 - msk] % mod;
        }
      }
      for (int sub_winners = (msk - 1) & msk; sub_winners > 0;
           sub_winners = (sub_winners - 1) & msk) {
        int sub_losers = msk ^ sub_winners;
        ll p = dp_prob[sub_winners];
        for (int bit = 0; bit < n; bit++) {
          if (sub_losers >> bit & 1) {
            p = p * win_msk[bit][(1 << n) - 1 - msk] % mod;
          }
        }
        dp_prob[msk] = (dp_prob[msk] - p + mod) % mod;
      }
      ans = (ans + dp_prob[msk] * num_winners % mod) % mod;
    }
    cout << ans << "\n";
    return 0;
  }
  