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 MOD = 1000000007;
  vector<int> pow_3(16, 1);
  static inline int weigth(int n) {
    if (n == 0) return 0;
    int res = 0;
    while (n) {
      int q = __builtin_ffs(n) - 1;
      n ^= (1 << q);
      res += pow_3[q];
    }
    return res;
  }
  int main(void) {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout << fixed << setprecision(14);
    vector<long long> inv(2000003, 1);
    for (int i = 2; i < 2000003; i++) {
      inv[i] = inv[MOD % i] * (MOD - MOD / i) % MOD;
    }
    for (int i = 1; i < 16; i++) pow_3[i] = 3 * pow_3[i - 1];
    int n;
    cin >> n;
    int N = 1 << n;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    vector<long long> masks(n, 1);
    for (int i = 1; i < n; i++) masks[i] = 1 << i;
    vector<vector<long long>> defeat_prob(n, vector<long long>(N, 1));
    for (int i = 0; i < n; i++) {
      for (int j = 1; j < N; j++) {
        if (!(j & masks[i])) {
          int least_sign_j = __builtin_ffs(j) - 1;
          defeat_prob[i][j] =
              (defeat_prob[i][j & (j - 1)] * a[least_sign_j]) % MOD;
          defeat_prob[i][j] =
              (defeat_prob[i][j] * inv[a[least_sign_j] + a[i]]) % MOD;
        }
      }
    }
    int length = 1;
    for (int i = 0; i < n; i++) length *= 3;
    vector<long long> probs(length, 1);
    vector<long long> win_prob(N, 0);
    for (int m = 0; m < N; m++) {
      int w_m = weigth(m);
      for (int s = m; s; s = (s - 1) & m) {
        int x = m ^ s;
        int w_x = weigth(x);
        int w_s = weigth(s);
        if (s == m) {
          probs[w_s] = 0;
          continue;
        }
        int ss = s;
        long long prob = 1;
        while (ss) {
          int q = __builtin_ffs(ss) - 1;
          prob = (prob * defeat_prob[q][x]) % MOD;
          ss ^= masks[q];
        }
        probs[2 * w_x + w_s] = (probs[2 * w_x] * prob) % MOD;
        probs[2 * w_m] = (probs[2 * w_m] - probs[2 * w_x + w_s] + MOD) % MOD;
      }
    }
    long long expected = 0;
    for (int m = 0; m < N; m++) {
      int count = 0;
      int mm = m;
      while (mm) {
        int q = __builtin_ffs(mm) - 1;
        count++;
        mm ^= masks[q];
      }
      int s = weigth((N - 1) ^ m);
      expected = (expected + count * probs[2 * weigth(m) + s]) % MOD;
    }
    cout << expected << "\n";
  }
  