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>
  const int M = 15;
  const int MOD = 1e9 + 7;
  long long read() {
    long long x = 0, f = 1;
    char c;
    while ((c = getchar()) < '0' || c > '9') {
      if (c == '-') f = -1;
    }
    while (c >= '0' && c <= '9') {
      x = (x << 3) + (x << 1) + (c ^ 48);
      c = getchar();
    }
    return x * f;
  }
  long long n, ans, a[M], b[1 << M], zs[M][1 << M], dp[1 << M], cnt[1 << M];
  long long qkpow(long long a, long long b) {
    long long r = 1;
    while (b > 0) {
      if (b & 1) r = r * a % MOD;
      a = a * a % MOD;
      b >>= 1;
    }
    return r;
  }
  signed main() {
    n = read();
    for (long long i = 0; i < n; i++) a[i] = read(), b[1 << i] = a[i];
    for (long long i = 0; i < n; i++) {
      zs[i][0] = 1;
      for (long long j = 1; j < (1 << n); j++) {
        long long x = j & (-j);
        zs[i][j] = zs[i][j ^ x] * a[i] % MOD * qkpow(a[i] + b[x], MOD - 2) % MOD;
      }
    }
    for (long long s = 1; s < (1 << n); s++) {
      dp[s] = 1;
      cnt[s] = cnt[s >> 1] + (s & 1);
      for (long long i = (s - 1) & s; i; i = (i - 1) & s) {
        long long p = 1;
        for (long long j = 0; j < n; j++)
          if (i & (1 << j)) p = p * zs[j][s - i] % MOD;
        dp[s] -= dp[i] * p;
        dp[s] %= MOD;
      }
    }
    for (long long s = 1; s < (1 << n); s++) {
      long long p = 1;
      for (long long i = 0; i < n; i++)
        if (s & (1 << i)) p = p * zs[i][(1 << n) - 1 - s] % MOD;
      ans += dp[s] * cnt[s] % MOD * p;
      ans %= MOD;
    }
    printf("%lld\n", (ans + MOD) % MOD);
  }
  