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 long long MOD = 1000000007LL;
  long long FastPow(long long k, long long p) {
    if (p == 0) return 1LL;
    long long P = FastPow(k, p / 2LL);
    P = (P * P) % MOD;
    if (p % 2LL == 1LL) {
      P = (P * k) % MOD;
    }
    return P;
  }
  long long Inv(long long x) { return FastPow(x, MOD - 2LL); }
  int n;
  long long a[21];
  long long probs[21][21];
  inline bool bit(int k, int b) { return (k & (1 << b)) > 0; }
  string tob(int k) {
    if (k == 0) return "0";
    string s;
    while (k > 0) {
      s.push_back(k % 2 + '0');
      k /= 2;
    }
    reverse(s.begin(), s.end());
    return s;
  }
  long long F[100111];
  long long caseProbPrec[15][100111];
  int main() {
    int i, j, in, ja;
    scanf("%d", &n);
    for (i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
    }
    long long ans = 0LL;
    for (int iter = 0; iter < n; iter++) {
      for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
          probs[i][j] = (a[i] * Inv(a[i] + a[j])) % MOD;
        }
      }
      for (i = 0; i < n - 1; i++) {
        for (int sub = 0; sub < (1 << (n - 1)); sub++) {
          caseProbPrec[i][sub] = probs[i][n - 1];
          for (j = 0; j < n - 1; j++) {
            if (bit(sub, j)) {
              caseProbPrec[i][sub] *= probs[i][j];
              caseProbPrec[i][sub] %= MOD;
            }
          }
        }
      }
      F[0] = 1LL;
      for (int mask = 1; mask < (1 << (n - 1)); mask++) {
        F[mask] = 1LL;
        for (int sub = ((mask - 1) & mask); sub >= 0; sub = (sub - 1) & mask) {
          long long caseProb = F[sub];
          for (i = 0; i < n - 1; i++) {
            if (bit(mask, i) && !bit(sub, i)) {
              caseProb *= caseProbPrec[i][sub];
              caseProb %= MOD;
            }
          }
          F[mask] -= caseProb;
          if (F[mask] < 0) F[mask] += MOD;
          if (sub == 0) break;
        }
      }
      ans += F[(1 << (n - 1)) - 1];
      if (ans >= MOD) ans -= MOD;
      swap(a[iter], a[n - 1]);
    }
    printf("%lld\n", ans);
    return 0;
  }
  