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 = (int)1e9 + 7;
  class Mint {
   private:
    int value;
  
   public:
    constexpr Mint() : value() {}
    template <typename T>
    Mint(const T& x) {
      value = (-mod <= x && x < mod) ? x : x % mod;
      if (value < 0) value += mod;
    }
    template <typename U>
    explicit operator U() const {
      return static_cast<U>(value);
    }
    Mint& operator+=(const Mint& a) {
      if ((value += a.value) >= mod) value -= mod;
      return *this;
    }
    Mint& operator-=(const Mint& a) {
      if ((value -= a.value) < 0) value += mod;
      return *this;
    }
    Mint& operator++() { return *this += 1; }
    Mint& operator--() { return *this -= 1; }
    Mint operator++(int) {
      Mint result(*this);
      *this += 1;
      return result;
    }
    Mint operator--(int) {
      Mint result(*this);
      *this -= 1;
      return result;
    }
    Mint operator-() const { return Mint(-value); }
    Mint& operator*=(const Mint& a) {
      value = (long long)value * a.value % mod;
      return *this;
    }
    Mint& operator/=(const Mint& a) {
      assert(a.value != 0);
      return *this *= inverse(a);
    }
    friend Mint pow(Mint a, int b) {
      Mint r = 1;
      for (; b; a *= a, b >>= 1) {
        if (b & 1) r *= a;
      }
      return r;
    }
    static Mint inverse(const Mint& a) { return pow(a, mod - 2); }
    friend bool operator==(const Mint& a, const Mint& b) {
      return a.value == b.value;
    }
    friend bool operator!=(const Mint& a, const Mint& b) {
      return a.value != b.value;
    }
    friend bool operator<(const Mint& a, const Mint& b) {
      return a.value < b.value;
    }
    friend Mint operator+(const Mint& a, const Mint& b) { return Mint(a) += b; }
    friend Mint operator-(const Mint& a, const Mint& b) { return Mint(a) -= b; }
    friend Mint operator*(const Mint& a, const Mint& b) { return Mint(a) *= b; }
    friend Mint operator/(const Mint& a, const Mint& b) { return Mint(a) /= b; }
    friend ostream& operator<<(ostream& stream, const Mint& a) {
      return stream << a.value;
    }
    friend istream& operator>>(istream& stream, Mint& a) {
      return stream >> a.value;
    }
    friend string to_string(const Mint& a) { return to_string(a.value); }
  };
  void foo() {
    int n;
    cin >> n;
    const int S = 1 << n;
    vector<int> f(n);
    for (int i = 0; i < n; i++) {
      cin >> f[i];
    }
    vector<vector<Mint>> p(n, vector<Mint>(n));
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        p[i][j] = Mint(f[i]) / (f[i] + f[j]);
      }
    }
    vector<vector<Mint>> all_out(n, vector<Mint>(S, 1));
    for (int i = 0; i < n; i++) {
      for (int s = 0; s < S; s++) {
        for (int u = 0; u < n; u++) {
          if (s & (1 << u)) {
            all_out[i][s] *= p[i][u];
          }
        }
      }
    }
    vector<map<int, Mint>> g(S);
    for (int s = 1; s < S; s++) {
      for (int j = s; j > 0; j = (j - 1) & s) {
        g[s][j] = 1;
        int other = s ^ j;
        for (int v = 0; v < n; v++) {
          if (j & (1 << v)) {
            g[s][j] *= all_out[v][other];
          }
        }
      }
    }
    vector<Mint> scc(S);
    for (int s = 1; s < S; s++) {
      scc[s] = 1;
      for (int sub = ((s - 1) & s); sub > 0; sub = (sub - 1) & s) {
        scc[s] -= scc[sub] * g[s][sub];
      }
    }
    Mint ans = 0;
    for (int i = 0; i < S; i++) {
      ans += scc[i] * g[S - 1][i] * __builtin_popcount(i);
    }
    cout << ans << endl;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    foo();
  }
  