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 T>
  T inverse(T a, T m) {
    T u = 0, v = 1;
    while (a != 0) {
      T t = m / a;
      m -= t * a;
      swap(a, m);
      u -= t * v;
      swap(u, v);
    }
    assert(m == 1);
    return u;
  }
  template <typename T>
  class Modular {
   public:
    using Type = typename decay<decltype(T::value)>::type;
    constexpr Modular() : value() {}
    template <typename U>
    Modular(const U& x) {
      value = normalize(x);
    }
    template <typename U>
    static Type normalize(const U& x) {
      Type v;
      if (-mod() <= x && x < mod())
        v = static_cast<Type>(x);
      else
        v = static_cast<Type>(x % mod());
      if (v < 0) v += mod();
      return v;
    }
    const Type& operator()() const { return value; }
    template <typename U>
    explicit operator U() const {
      return static_cast<U>(value);
    }
    constexpr static Type mod() { return T::value; }
    Modular& operator+=(const Modular& other) {
      if ((value += other.value) >= mod()) value -= mod();
      return *this;
    }
    Modular& operator-=(const Modular& other) {
      if ((value -= other.value) < 0) value += mod();
      return *this;
    }
    template <typename U>
    Modular& operator+=(const U& other) {
      return *this += Modular(other);
    }
    template <typename U>
    Modular& operator-=(const U& other) {
      return *this -= Modular(other);
    }
    Modular& operator++() { return *this += 1; }
    Modular& operator--() { return *this -= 1; }
    Modular operator++(int) {
      Modular result(*this);
      *this += 1;
      return result;
    }
    Modular operator--(int) {
      Modular result(*this);
      *this -= 1;
      return result;
    }
    Modular operator-() const { return Modular(-value); }
    template <typename U = T>
    typename enable_if<is_same<typename Modular<U>::Type, int>::value,
                       Modular>::type&
    operator*=(const Modular& rhs) {
      value = normalize(static_cast<int64_t>(value) *
                        static_cast<int64_t>(rhs.value));
      return *this;
    }
    template <typename U = T>
    typename enable_if<is_same<typename Modular<U>::Type, int64_t>::value,
                       Modular>::type&
    operator*=(const Modular& rhs) {
      int64_t q = static_cast<int64_t>(static_cast<long double>(value) *
                                       rhs.value / mod());
      value = normalize(value * rhs.value - q * mod());
      return *this;
    }
    template <typename U = T>
    typename enable_if<!is_integral<typename Modular<U>::Type>::value,
                       Modular>::type&
    operator*=(const Modular& rhs) {
      value = normalize(value * rhs.value);
      return *this;
    }
    Modular& operator/=(const Modular& other) {
      return *this *= Modular(inverse(other.value, mod()));
    }
    template <typename U>
    friend const Modular<U>& abs(const Modular<U>& v) {
      return v;
    }
    template <typename U>
    friend bool operator==(const Modular<U>& lhs, const Modular<U>& rhs);
    template <typename U>
    friend bool operator<(const Modular<U>& lhs, const Modular<U>& rhs);
    template <typename U>
    friend std::istream& operator>>(std::istream& stream, Modular<U>& number);
  
   private:
    Type value;
  };
  template <typename T>
  bool operator==(const Modular<T>& lhs, const Modular<T>& rhs) {
    return lhs.value == rhs.value;
  }
  template <typename T, typename U>
  bool operator==(const Modular<T>& lhs, U rhs) {
    return lhs == Modular<T>(rhs);
  }
  template <typename T, typename U>
  bool operator==(U lhs, const Modular<T>& rhs) {
    return Modular<T>(lhs) == rhs;
  }
  template <typename T>
  bool operator!=(const Modular<T>& lhs, const Modular<T>& rhs) {
    return !(lhs == rhs);
  }
  template <typename T, typename U>
  bool operator!=(const Modular<T>& lhs, U rhs) {
    return !(lhs == rhs);
  }
  template <typename T, typename U>
  bool operator!=(U lhs, const Modular<T>& rhs) {
    return !(lhs == rhs);
  }
  template <typename T>
  bool operator<(const Modular<T>& lhs, const Modular<T>& rhs) {
    return lhs.value < rhs.value;
  }
  template <typename T>
  Modular<T> operator+(const Modular<T>& lhs, const Modular<T>& rhs) {
    return Modular<T>(lhs) += rhs;
  }
  template <typename T, typename U>
  Modular<T> operator+(const Modular<T>& lhs, U rhs) {
    return Modular<T>(lhs) += rhs;
  }
  template <typename T, typename U>
  Modular<T> operator+(U lhs, const Modular<T>& rhs) {
    return Modular<T>(lhs) += rhs;
  }
  template <typename T>
  Modular<T> operator-(const Modular<T>& lhs, const Modular<T>& rhs) {
    return Modular<T>(lhs) -= rhs;
  }
  template <typename T, typename U>
  Modular<T> operator-(const Modular<T>& lhs, U rhs) {
    return Modular<T>(lhs) -= rhs;
  }
  template <typename T, typename U>
  Modular<T> operator-(U lhs, const Modular<T>& rhs) {
    return Modular<T>(lhs) -= rhs;
  }
  template <typename T>
  Modular<T> operator*(const Modular<T>& lhs, const Modular<T>& rhs) {
    return Modular<T>(lhs) *= rhs;
  }
  template <typename T, typename U>
  Modular<T> operator*(const Modular<T>& lhs, U rhs) {
    return Modular<T>(lhs) *= rhs;
  }
  template <typename T, typename U>
  Modular<T> operator*(U lhs, const Modular<T>& rhs) {
    return Modular<T>(lhs) *= rhs;
  }
  template <typename T>
  Modular<T> operator/(const Modular<T>& lhs, const Modular<T>& rhs) {
    return Modular<T>(lhs) /= rhs;
  }
  template <typename T, typename U>
  Modular<T> operator/(const Modular<T>& lhs, U rhs) {
    return Modular<T>(lhs) /= rhs;
  }
  template <typename T, typename U>
  Modular<T> operator/(U lhs, const Modular<T>& rhs) {
    return Modular<T>(lhs) /= rhs;
  }
  template <typename T, typename U>
  Modular<T> power(const Modular<T>& a, const U& b) {
    assert(b >= 0);
    Modular<T> x = a, res = 1;
    U p = b;
    while (p > 0) {
      if (p & 1) res *= x;
      x *= x;
      p >>= 1;
    }
    return res;
  }
  template <typename T>
  bool IsZero(const Modular<T>& number) {
    return number() == 0;
  }
  template <typename T>
  string to_string(const Modular<T>& number) {
    return to_string(number());
  }
  template <typename T>
  std::ostream& operator<<(std::ostream& stream, const Modular<T>& number) {
    return stream << number();
  }
  template <typename T>
  std::istream& operator>>(std::istream& stream, Modular<T>& number) {
    typename common_type<typename Modular<T>::Type, int64_t>::type x;
    stream >> x;
    number.value = Modular<T>::normalize(x);
    return stream;
  }
  constexpr int md = (int)1e9 + 7;
  using Mint = Modular<std::integral_constant<decay<decltype(md)>::type, md>>;
  int n;
  Mint a[14];
  Mint adj[14][14];
  const int nax = 5e6;
  int cached[nax];
  Mint mem[nax];
  Mint prob[2][14][1 << 14];
  int ind3[1 << 14];
  Mint f(int done, int reach, int left) {
    if (left == 0) return 1;
    if (reach == 0) return 0;
    int memi = ind3[reach] * 2 + ind3[left];
    Mint& memo = mem[memi];
    if (!cached[memi]++) {
      int p = 0;
      while (!(reach >> p & 1)) p++;
      Mint ans = 0;
      int m = left;
      do {
        ans += prob[1][p][left & m] * prob[0][p][left & ~m] *
               f(done | (1 << p), (reach ^ (1 << p)) | m, left ^ m);
        m = (m - 1) & left;
      } while (m != left);
      memo = ans;
    }
    return memo;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int x = 0; x < 1 << n; x++) {
      for (int i = 0; i < n; i++) {
        ind3[x] = ind3[x] * 3 + (x >> i & 1);
      }
    }
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) adj[i][j] = a[i] / (a[i] + a[j]);
    for (int k : {0, 1}) {
      for (int p = 0; p < n; p++) {
        for (int left = 0; left < 1 << n; left++) {
          prob[k][p][left] = 1;
          for (int i = 0; i < n; i++) {
            if (!(left >> i & 1)) continue;
            prob[k][p][left] *= (k ? adj[p][i] : 1 - adj[p][i]);
          }
        }
      }
    }
    Mint ans = 0;
    for (int i = 0; i < n; i++) {
      ans += f(0, 1 << i, ((1 << n) - 1) ^ (1 << i));
    }
    cout << ans << endl;
  }
  