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;
  const int N = 15;
  const int M = 1 << N;
  int a[N], dp[M], p[N][M];
  int add(int a, int b) { return (a + b) % mod; }
  int mul(int a, int b) { return 1ll * a * b % mod; }
  int qpow(int a, int n) {
    int re = 1;
    while (n > 0) {
      if (n & 1) {
        re = mul(re, a);
      }
      a = mul(a, a);
      n >>= 1;
    }
    return re;
  }
  int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    for (int i = 0; i < (1 << n); i++) {
      for (int j = 0; j < n; j++) {
        p[j][i] = 1;
      }
      for (int j = 0; j < n; j++) {
        if (i >> j & 1) {
          for (int k = 0; k < n; k++) {
            if (i >> k & 1) continue;
            p[k][i] =
                mul(p[k][i ^ 1 << j], mul(a[k], qpow(add(a[k], a[j]), mod - 2)));
          }
          break;
        }
      }
    }
    int ans = 0;
    for (int i = 1; i < (1 << n); i++) {
      dp[i] = 1;
      for (int j = (i - 1) & i; j; j = (j - 1) & i) {
        int tmp = 1;
        for (int k = 0; k < n; k++) {
          if (j >> k & 1) {
            tmp = mul(tmp, p[k][i ^ j]);
          }
        }
        dp[i] = add(dp[i], mod - mul(dp[j], tmp));
      }
      int tmp = 1;
      for (int j = 0; j < n; j++) {
        if (i >> j & 1) {
          tmp = mul(tmp, p[j][(1 << n) - 1 ^ i]);
        }
      }
      ans = add(ans, mul(mul(dp[i], tmp), __builtin_popcount(i)));
    }
    printf("%d", ans);
    return 0;
  }
  