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(3)
  #pragma GCC optimize(2)
  using namespace std;
  const int maxv = 1e6 + 10;
  const int maxn = 2e5 + 10;
  const int inf32 = 1e9 + 10;
  const long long inf64 = 1e18 + 10;
  const int mod = 1e9 + 7;
  long long mul(long long first, long long second) {
    return first * second % mod;
  }
  long long add(long long first, long long second) {
    return (first + second) % mod;
  }
  long long sub(long long first, long long second) {
    return ((first - second) % mod + mod) % mod;
  }
  long long qpow(long long first, long long n) {
    long long ans = 1;
    while (n) {
      if (n & 1) ans = mul(ans, first);
      first = mul(first, first);
      n /= 2;
    }
    return ans;
  }
  long long inverse(long long first) { return qpow(first, mod - 2); }
  int N, a[14];
  long long inv[maxv * 2], g[14][1 << 14], dp[1 << 14];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    cin >> N;
    for (int i = 0; i < N; ++i) cin >> a[i];
    for (int i = 1; i < maxv * 2; ++i) inv[i] = inverse(i);
    for (int i = 0; i < N; ++i) {
      for (int mask = 0; mask < 1 << N; ++mask) {
        if (mask >> i & 1) continue;
        g[i][mask] = 1;
        for (int j = 0; j < N; ++j) {
          if (mask >> j & 1) {
            g[i][mask] = mul(g[i][mask], mul(a[i], inv[a[i] + a[j]]));
          }
        }
      }
    }
    for (int mask = 1; mask < 1 << N; ++mask) {
      dp[mask] = 1;
      for (int submask = mask - 1; submask > 0; submask = (submask - 1) & mask) {
        long long v = 1;
        for (int i = 0; i < N; ++i) {
          if (submask >> i & 1) {
            v = mul(v, g[i][mask ^ submask]);
          }
        }
        dp[mask] = sub(dp[mask], mul(dp[submask], v));
      }
    }
    long long ans = 0;
    for (int mask = 1; mask < 1 << N; ++mask) {
      long long v = 1;
      for (int i = 0; i < N; ++i) {
        if (mask >> i & 1) {
          v = mul(v, g[i][((1 << N) - 1) ^ mask]);
        }
      }
      ans = add(ans, mul(mul(dp[mask], v), __builtin_popcount(mask)));
    }
    cout << ans << endl;
  }
  