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;
  inline int add(int x, int y) {
    x += y;
    if (x >= mod) x -= mod;
    return x;
  }
  inline int sub(int x, int y) {
    x -= y;
    if (x < 0) x += mod;
    return x;
  }
  inline int mul(int x, int y) { return (x * 1ll * y) % mod; }
  inline int powr(int a, long long b) {
    int x = 1 % mod;
    while (b) {
      if (b & 1) x = mul(x, a);
      a = mul(a, a);
      b >>= 1;
    }
    return x;
  }
  inline int inv(int a) { return powr(a, mod - 2); }
  int p[20][20];
  int a[20];
  int probSCC[1 << 14];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      for (int j = 0; j < i; j++) {
        p[i][j] = mul(a[i], inv(a[i] + a[j]));
        p[j][i] = mul(a[j], inv(a[i] + a[j]));
      }
    }
    for (int i = 1; i < (1 << n); i++) {
      probSCC[i] = 1;
      for (int j = i; j; j = (j - 1) & i) {
        if (j == i) continue;
        int prob = 1;
        int _j = i ^ j;
        for (int u = 0; u < n; u++)
          if (j >> u & 1) {
            for (int v = 0; v < n; v++)
              if (_j >> v & 1) {
                prob = mul(prob, p[u][v]);
              }
          }
        probSCC[i] = sub(probSCC[i], mul(probSCC[j], prob));
      }
    }
    int ans = 0;
    int i = (1 << n) - 1;
    for (int j = i; j; j = (j - 1) & i) {
      int prob = probSCC[j];
      int _j = i ^ j;
      for (int u = 0; u < n; u++)
        if (j >> u & 1) {
          for (int v = 0; v < n; v++)
            if (_j >> v & 1) {
              prob = mul(prob, p[u][v]);
            }
        }
      ans = add(ans, mul(__builtin_popcount(j), prob));
    }
    cout << ans << '\n';
  }
  