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;
  long long fpow(long long a, long long b) {
    long long r = 1;
    for (a %= mod; b; b >>= 1) {
      if (b & 1) r = r * a % mod;
      a = a * a % mod;
    }
    return r;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    long long a[n];
    for (int i = 0; i < n; ++i) cin >> a[i];
    long long win[n][n];
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (i == j) continue;
        win[i][j] = a[i] * fpow(a[i] + a[j], mod - 2) % mod;
      }
    }
    long long dp[1 << n], mul[1 << n][n], inner[1 << n][2];
    for (int i = 1; i < 1 << n; ++i) {
      mul[i][0] = mul[i][1] = 1;
      for (int j = 0; j < n; ++j) {
        if (i >> j & 1) mul[i][1] = mul[i][1] * a[j] % mod;
      }
      for (int j = 2; j < n; ++j) {
        mul[i][j] = mul[i][j - 1] * mul[i][1] % mod;
      }
      inner[i][0] = 1;
      for (int j = 0; j < n; ++j) {
        if (i >> j & 1)
          for (int k = j + 1; k < n; ++k) {
            if (i >> k & 1) inner[i][0] = inner[i][0] * (a[j] + a[k]) % mod;
          }
      }
      inner[i][1] = fpow(inner[i][0], mod - 2);
    }
    for (int i = 1; i < 1 << n; ++i) {
      dp[i] = 1;
      for (int j = (i - 1) & i; j; j = (j - 1) & i) {
        dp[i] -= dp[j] * mul[j][__builtin_popcount(i ^ j)] % mod * inner[j][0] %
                 mod * inner[i ^ j][0] % mod * inner[i][1] % mod;
      }
      dp[i] %= mod;
    }
    long long ans = 0;
    for (int i = 1; i < 1 << n; ++i) {
      long long tmp = 1;
      for (int j = 0; j < n; ++j) {
        if (i >> j & 1)
          for (int k = 0; k < n; ++k) {
            if (!(i >> k & 1)) tmp = tmp * win[j][k] % mod;
          }
      }
      ans += tmp * dp[i] % mod * __builtin_popcount(i) % mod;
    }
    ans %= mod;
    if (ans < 0) ans += mod;
    cout << ans << '\n';
  }
  