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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <int MOD>
  struct Modular {
    int val;
    Modular() : val(0){};
    Modular(int _val) : val(_val % MOD){};
    void norm() {
      val %= MOD;
      if (val < 0) val += MOD;
    }
    Modular& operator+=(const Modular& other) {
      val += other.val;
      if (val >= MOD) val -= MOD;
      return *this;
    }
    Modular operator+(const Modular& other) const {
      auto tmp(*this);
      tmp += other;
      return tmp;
    }
    Modular& operator-=(const Modular& other) {
      val -= other.val;
      if (val < 0) val += MOD;
      return *this;
    }
    Modular operator-(const Modular& other) const {
      auto tmp(*this);
      tmp -= other;
      return tmp;
    }
    Modular& operator*=(const Modular& other) {
      val = (val * 1LL * other.val) % MOD;
      return *this;
    }
    Modular operator*(const Modular& other) const {
      auto tmp(*this);
      tmp *= other;
      return tmp;
    }
    Modular bin_pow(int n) const {
      Modular res(1);
      Modular a = *this;
      for (; n; n >>= 1, a *= a)
        if (n & 1) res *= a;
      return res;
    }
    Modular inv() const { return this->bin_pow(MOD - 2); }
    Modular& operator/=(const Modular& other) { return *this *= other.inv(); }
    Modular operator/(const Modular& other) const { return *this * other.inv(); }
  };
  template <int MOD>
  std::ostream& operator<<(std::ostream& out, const Modular<MOD>& x) {
    out << x.val;
    return out;
  }
  const int MOD = (int)1e9 + 7;
  using Mint = Modular<MOD>;
  int bit(int mask, int k) { return (mask >> k) & 1; }
  const int N = 14;
  int a[N];
  Mint p[N][1 << N];
  Mint scc[1 << N];
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
    }
    auto getp = [&](int i, int j) { return Mint(a[i]) / Mint(a[i] + a[j]); };
    auto getpp = [&](int mask1, int mask2) {
      Mint res(1);
      for (int i = 0; i < n; ++i) {
        if (bit(mask1, i)) {
          res *= p[i][mask2];
        }
      }
      return res;
    };
    for (int i = 0; i < n; ++i) {
      for (int mask = 0; mask < (1 << n); ++mask) {
        if (bit(mask, i)) continue;
        p[i][mask] = Mint(1);
        for (int j = 0; j < n; ++j) {
          if (bit(mask, j)) {
            p[i][mask] *= getp(i, j);
          }
        }
      }
    }
    Mint ans(0);
    for (int mask = 0; mask < (1 << n); ++mask) {
      scc[mask] = Mint(1);
      if (__builtin_popcount(mask) > 1) {
        for (int smask = (mask - 1) & mask; smask; smask = (smask - 1) & mask) {
          scc[mask] -= scc[smask] * getpp(smask, mask ^ smask);
        }
      }
      ans += Mint(__builtin_popcount(mask)) * scc[mask] *
             getpp(mask, mask ^ ((1 << n) - 1));
    }
    cout << ans << '\n';
  }
  