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;
  long long n, m, k;
  long long read() {
    long long s = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      s = s * 10 + ch - '0';
      ch = getchar();
    }
    return s * f;
  }
  long long a[15];
  long long dp[(1 << 14) + 1];
  long long b1[20], b2[20];
  long long inv[2020020];
  long long t1, t2;
  const long long mod = 1e9 + 7;
  signed main() {
    long long T = 1;
    inv[1] = 1;
    inv[0] = 1;
    for (long long i = 2, iend = 2000000; i <= iend; ++i)
      inv[i] = 1ll * inv[mod % i] * (mod - mod / i) % mod;
    while (T--) {
      n = read();
      for (long long i = 0, iend = n - 1; i <= iend; ++i) a[i] = read();
      long long m = (1 << n) - 1;
      for (long long i = 1, iend = m; i <= iend; ++i) {
        dp[i] = 1;
        for (long long j = (i - 1) & i; j; j = (j - 1) & i) {
          long long k = i - j;
          t1 = t2 = 0;
          for (long long p = 0, pend = n - 1; p <= pend; ++p) {
            if (j & (1 << p)) {
              b1[++t1] = p;
            }
            if (k & (1 << p)) {
              b2[++t2] = p;
            }
          }
          long long nw = dp[j];
          for (long long f1 = 1, f1end = t1; f1 <= f1end; ++f1) {
            for (long long f2 = 1, f2end = t2; f2 <= f2end; ++f2) {
              nw *= a[b1[f1]];
              nw %= mod;
              nw *= inv[a[b1[f1]] + a[b2[f2]]];
              nw %= mod;
            }
          }
          dp[i] += mod, dp[i] -= nw, dp[i] %= mod;
        }
      }
      long long sum = 0;
      for (long long i = 1, iend = m; i <= iend; ++i) {
        t1 = t2 = 0;
        for (long long p = 0, pend = n - 1; p <= pend; ++p) {
          if (i & (1 << p)) {
            b1[++t1] = p;
          }
          if ((m ^ i) & (1 << p)) {
            b2[++t2] = p;
          }
        }
        long long nw = dp[i];
        for (long long f1 = 1, f1end = t1; f1 <= f1end; ++f1) {
          for (long long f2 = 1, f2end = t2; f2 <= f2end; ++f2) {
            nw *= a[b1[f1]];
            nw %= mod;
            nw *= inv[a[b1[f1]] + a[b2[f2]]];
            nw %= mod;
          }
        }
        nw *= t1;
        nw %= mod;
        sum += nw;
        sum %= mod;
      }
      cout << sum;
    }
  }
  