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 N = 14 + 1;
  const int mod = 1e9 + 7;
  int sum(int a, int b) {
    int c = a + b;
    if (c >= mod) c -= mod;
    return c;
  }
  int mul(int a, int b) { return 1ll * a * b % mod; }
  int bpow(int a, int b) {
    int r = 1;
    for (; b; b /= 2) {
      if (b % 2) r = mul(r, a);
      a = mul(a, a);
    }
    return r;
  }
  int dp[N][(1 << N)];
  int a[N];
  int win[N][N];
  int wmask[N][(1 << N)];
  int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++)
        if (i != j) win[i][j] = mul(a[i], bpow(sum(a[i], a[j]), mod - 2));
    }
    for (int i = 0; i < n; i++) {
      wmask[i][0] = 1;
      for (int mask = 1; mask < (1 << n); mask++) {
        int j = __builtin_ctz(mask);
        wmask[i][mask] = mul(win[j][i], wmask[i][mask ^ (1 << j)]);
      }
    }
    for (int mask = 1; mask < (1 << n); mask++) {
      for (int i = 0; i < n; i++)
        if ((mask >> i) & 1) dp[i][mask] = 1;
      for (int bmask = ((mask - 1) & mask); bmask; bmask = ((bmask - 1) & mask)) {
        int prob = 1;
        for (int i = 0; i < n; i++)
          if ((bmask >> i) & 1) prob = mul(prob, wmask[i][bmask ^ mask]);
        for (int i = 0; i < n; i++)
          if ((bmask >> i) & 1)
            dp[i][mask] = sum(dp[i][mask], mod - mul(prob, dp[i][bmask]));
      }
    }
    int ans = 0;
    for (int i = 0; i < n; i++) ans = sum(ans, dp[i][(1 << n) - 1]);
    cout << ans << endl;
    return 0;
  }
  