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;
  int n, Mx;
  long long res, ans = 0;
  int a[16], one[16386];
  long long f[16386], fac[2000002], inv[2000002];
  template <class T>
  void read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9')
      x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
    x = f ? -x : x;
    return;
  }
  inline long long Pow(long long a, int b) {
    if (!b) return 1;
    if (b == 1) return a;
    long long c = Pow(a, (b >> 1));
    c = (c * c) % 1000000007;
    if (b & 1) c = (c * a) % 1000000007;
    return c;
  }
  inline void init() {
    fac[0] = 1;
    for (int i = 1; i <= 2000000; ++i) fac[i] = (fac[i - 1] * i) % 1000000007;
    inv[2000000] = Pow(fac[2000000], 1000000007 - 2);
    for (int i = 2000000; i; --i)
      inv[i - 1] = (inv[i] * i) % 1000000007,
              inv[i] = (inv[i] * fac[i - 1]) % 1000000007;
  }
  int main() {
    init(), read(n), Mx = (1 << n);
    for (int i = 0; i < n; ++i) read(a[i]);
    for (int i = 1; i < Mx; ++i) {
      one[i] = one[(i >> 1)] + (i & 1);
      for (int S = (i - 1) & i; S; S = (S - 1) & i) {
        res = 1 - f[S];
        for (int j = 0; j < n; ++j)
          if ((S >> j) & 1)
            for (int k = 0; k < n; ++k)
              if (((i >> k) & 1) && !((S >> k) & 1))
                res =
                    (((res * a[j]) % 1000000007) * inv[a[j] + a[k]]) % 1000000007;
        f[i] = (f[i] + res) % 1000000007;
      }
      res = ((1LL - f[i]) * one[i]) % 1000000007;
      for (int j = 0; j < n; ++j)
        if ((i >> j) & 1)
          for (int k = 0; k < n; ++k)
            if (!((i >> k) & 1))
              res = (((res * a[j]) % 1000000007) * inv[a[j] + a[k]]) % 1000000007;
      ans = (ans + res) % 1000000007;
    }
    return 0 & printf("%lld", (ans + 1000000007) % 1000000007);
  }
  