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;
  template <typename P, typename Q>
  P CeilDiv(P _dividend, Q _divider) {
    return static_cast<P>((_dividend - 1LL + _divider) / _divider);
  }
  template <typename Tp>
  inline void outarr(Tp _begin, Tp _end, const char* _delim = " ") {
    for (Tp current = _begin; current != _end; ++current) {
      std::cout << *current << _delim;
    }
    std::cout << '\n';
  }
  using ll = long long;
  using pii = std::pair<int, int>;
  constexpr int INF = 0x3f3f3f3f;
  constexpr int MOD = static_cast<int>(1e9 + 7);
  int Add(int x, int y) {
    x += y;
    return x < MOD ? x : x - MOD;
  }
  int Sub(int x, int y) {
    x -= y;
    return x >= 0 ? x : x + MOD;
  }
  int Mul(int x, int y) { return int(1LL * x * y % MOD); }
  int FastPow(int x, int p) {
    int res = 1;
    while (p > 0) {
      if (p & 1) {
        res = Mul(res, x);
      }
      p >>= 1;
      x = Mul(x, x);
    }
    return res;
  }
  int Inv(int x) { return FastPow(x, MOD - 2); }
  bool HasBit(int x, int b) { return (x >> b) & 1; }
  void SolveTask() {
    int n;
    cin >> n;
    vector<int> arr(n);
    for (int i = 0; i < n; ++i) {
      cin >> arr[i];
    }
    vector<vector<int>> prob(n, vector<int>(n));
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        prob[i][j] = Mul(arr[i], Inv(arr[i] + arr[j]));
      }
    }
    const int full_mask = 1 << n;
    vector<int> dp(full_mask, 1);
    for (int mask = 1; mask < full_mask; ++mask) {
      for (int sub = (mask - 1) & mask; sub > 0; sub = (sub - 1) & mask) {
        int p = 1;
        for (int a = 0; a < n; ++a) {
          if (HasBit(sub, a)) {
            for (int b = 0; b < n; ++b) {
              if (HasBit(mask ^ sub, b)) {
                p = Mul(p, prob[a][b]);
              }
            }
          }
        }
        dp[mask] = Sub(dp[mask], Mul(dp[sub], p));
      }
    }
    vector<int> kbit(full_mask, 0);
    int ans = 0;
    for (int mask = 1; mask < full_mask; ++mask) {
      kbit[mask] = kbit[mask >> 1] + (mask & 1);
      for (int a = 0; a < n; ++a) {
        if (HasBit(mask, a)) {
          for (int b = 0; b < n; ++b) {
            if (!HasBit(mask, b)) {
              dp[mask] = Mul(dp[mask], prob[a][b]);
            }
          }
        }
      }
      ans = Add(ans, Mul(dp[mask], kbit[mask]));
    }
    cout << ans << "\n";
  }
  int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int test_cases = 1;
    for (int yt = 1; yt <= test_cases; ++yt) {
      SolveTask();
    }
    return 0;
  }
  