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 = 2e5 + 10, mod = 1e9 + 7;
  int pw(int x, int y) {
    int ret = 1;
    for (; y; y /= 2) {
      if (y & 1) {
        ret = 1ll * ret * x % mod;
      }
      x = 1ll * x * x % mod;
    }
    return ret;
  }
  void solve() {
    int n;
    cin >> n;
    int a[n], w[n][n];
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      for (int j = 0; j < i; j++) {
        w[i][j] = 1ll * a[i] * pw(a[i] + a[j], mod - 2) % mod;
        w[j][i] = (1 - w[i][j] + mod) % mod;
      }
    }
    int win[n][1 << n];
    for (int i = 0; i < n; i++) {
      win[i][0] = 1;
      for (int j = 1; j < (1 << n); j++) {
        if ((1 << i) & j) {
          continue;
        }
        win[i][j] = 1;
        for (int k = 0; k < n; k++) {
          if ((1 << k) & j) {
            win[i][j] = 1ll * win[i][j] * w[i][k] % mod;
          }
        }
      }
    }
    int dp[1 << n];
    int ans = 0;
    for (int i = 1; i < (1 << n); i++) {
      dp[i] = 1;
      for (int j = (i - 1) & i; j > 0; j = (j - 1) & i) {
        int val = dp[j];
        for (int k = 0; k < n; k++) {
          if ((1 << k) & j) {
            val = 1ll * val * win[k][i ^ j] % mod;
          }
        }
        dp[i] -= val;
        if (dp[i] < 0) {
          dp[i] += mod;
        }
      }
      int add = dp[i];
      for (int j = 0; j < n; j++) {
        if ((1 << j) & i) {
          add = 1ll * add * win[j][((1 << n) - 1) ^ i] % mod;
        }
      }
      ans += 1ll * add * __builtin_popcount(i) % mod;
      if (ans >= mod) {
        ans -= mod;
      }
    }
    cout << ans;
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    int cur = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  