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 mod = 1e9 + 7;
  int n, a[20], p[20][20], inv[2000010], lb[(1 << 14) + 10], cnt[(1 << 14) + 10],
      f[(1 << 14) + 10], s[20][(1 << 14) + 10], ns[(1 << 14) + 10],
      it[20][(1 << 14) + 10], v[(1 << 14) + 10], sz, ans;
  int POW(int x, int y) {
    int ret = 1;
    while (y)
      y& 1 ? ret = 1ll * ret * x % mod : 0, x = 1ll * x * x % mod, y >>= 1;
    return ret;
  }
  int main() {
    scanf("%d", &n), inv[1] = 1;
    for (int i = 2; i <= 2e6; ++i)
      inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
    for (int i = 1; i < (1 << n); ++i)
      lb[i] = (i & 1 ? 1 : lb[i >> 1] + 1), cnt[i] = cnt[i >> 1] + (i & 1);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
      for (int j = 1; j < i; ++j)
        p[i][j] = 1ll * a[i] * inv[a[i] + a[j]] % mod,
        (p[j][i] = (1 - p[i][j] + mod)) >= mod ? p[j][i] -= mod : 0;
    }
    for (int i = 1; i <= n; ++i) {
      s[i][0] = it[i][0] = 1;
      for (int j = 1; j < (1 << n); ++j)
        s[i][j] = 1ll * s[i][j - (j & -j)] * p[i][lb[j]] % mod,
        it[i][j] = 1ll * it[i][j - (j & -j)] * POW(p[lb[j]][i], mod - 2) % mod;
    }
    for (int i = 1; i < (1 << n); ++i) {
      f[i] = 1, sz = 0;
      for (int j = (i - 1 & i); j; j = (j - 1 & i)) v[++sz] = j;
      ns[0] = 1;
      for (int j = sz, t; j >= 1; --j) {
        t = v[j], ns[t] = 1ll * ns[t - (t & -t)] * s[lb[t]][i ^ t] % mod *
                          it[lb[t]][t - (t & -t)] % mod;
        (f[i] += mod - 1ll * f[t] * ns[t] % mod) >= mod ? f[i] -= mod : 0;
      }
    }
    ns[0] = 1;
    for (int i = 1; i < (1 << n); ++i) {
      ns[i] = 1ll * ns[i - (i & -i)] * s[lb[i]][(1 << n) - 1 ^ i] % mod *
              it[lb[i]][i - (i & -i)] % mod;
      ans = (ans + 1ll * f[i] * ns[i] % mod * cnt[i]) % mod;
    }
    printf("%d", ans);
    return 0;
  }
  