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;
  const int N = 1 << 14, mod = 1e9 + 7;
  int n, n1, n2, l1, l2, lim, a[14], e[14][14], f[4][128][128], dp[N], ans;
  void qmo(int &x) { x += x >> 31 & mod; }
  int ksm(int a, int b) {
    int res = 1;
    for (; b; b >>= 1, a = (long long)a * a % mod)
      if (b & 1) res = (long long)res * a % mod;
    return res;
  }
  int calc(int S, int T) {
    return (long long)f[0][S & l1 - 1][T & l1 - 1] * f[1][S & l1 - 1][T >> n1] %
           mod * f[2][S >> n1][T & l1 - 1] % mod * f[3][S >> n1][T >> n1] % mod;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin >> n;
    n1 = n >> 1;
    n2 = n + 1 >> 1;
    l1 = 1 << n1;
    l2 = 1 << n2;
    lim = 1 << n;
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      for (int j = 0; j < i; ++j) {
        e[i][j] = (long long)a[i] * ksm(a[i] + a[j], mod - 2) % mod;
        qmo(e[j][i] = mod + 1 - e[i][j]);
      }
    }
    for (int S = 0; S < l1; ++S)
      for (int T = 0; T < l1; ++T)
        if (!(S & T)) {
          f[0][S][T] = 1;
          for (int i = 0; i < n1; ++i)
            if (S >> i & 1)
              for (int j = 0; j < n1; ++j)
                if (T >> j & 1)
                  f[0][S][T] = (long long)f[0][S][T] * e[i][j] % mod;
        }
    for (int S = 0; S < l2; ++S)
      for (int T = 0; T < l2; ++T)
        if (!(S & T)) {
          f[3][S][T] = 1;
          for (int i = 0; i < n2; ++i)
            if (S >> i & 1)
              for (int j = 0; j < n2; ++j)
                if (T >> j & 1)
                  f[3][S][T] = (long long)f[3][S][T] * e[i + n1][j + n1] % mod;
        }
    for (int S = 0; S < l1; ++S)
      for (int T = 0; T < l2; ++T) {
        f[1][S][T] = 1;
        for (int i = 0; i < n1; ++i)
          if (S >> i & 1)
            for (int j = 0; j < n2; ++j)
              if (T >> j & 1)
                f[1][S][T] = (long long)f[1][S][T] * e[i][j + n1] % mod;
      }
    for (int S = 0; S < l2; ++S)
      for (int T = 0; T < l1; ++T) {
        f[2][S][T] = 1;
        for (int i = 0; i < n2; ++i)
          if (S >> i & 1)
            for (int j = 0; j < n1; ++j)
              if (T >> j & 1)
                f[2][S][T] = (long long)f[2][S][T] * e[i + n1][j] % mod;
      }
    for (int i = 1; i <= n; ++i) {
      for (int S = 1; S < lim; ++S)
        if (__builtin_popcount(S) == i) {
          dp[S] = 1;
          for (int T = S & S - 1; T; T = T - 1 & S)
            qmo(dp[S] -= (long long)dp[T] * calc(T, S - T) % mod);
          ans = (ans + (long long)dp[S] * i % mod * calc(S, lim - 1 - S)) % mod;
        }
    }
    printf("%d\n", ans);
  }
  