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 <class T, const T MOD>
  struct ModularInteger {
    T v;
    ModularInteger(T x = 0) {
      v = x % MOD;
      if (v < 0) {
        v += MOD;
      }
    }
    ModularInteger& operator+=(const ModularInteger& u) {
      v = (v + u.v) % MOD;
      return *this;
    }
    ModularInteger& operator-=(const ModularInteger& u) {
      v = ((v - u.v) % MOD + MOD) % MOD;
      return *this;
    }
    ModularInteger& operator*=(const ModularInteger& u) {
      v = v * u.v % MOD;
      return *this;
    }
    ModularInteger& operator/=(const ModularInteger& u) {
      return *this *= u.inverse();
    }
    ModularInteger operator-() const { return ModularInteger(-v); }
    ModularInteger operator+(const ModularInteger& u) const {
      return ModularInteger(v) += u.v;
    }
    ModularInteger operator-(const ModularInteger& u) const {
      return ModularInteger(v) -= u.v;
    }
    ModularInteger operator*(const ModularInteger& u) const {
      return ModularInteger(v) *= u.v;
    }
    ModularInteger operator/(const ModularInteger& u) const {
      return ModularInteger(v) /= u.v;
    }
    ModularInteger inverse() const {
      T a = v, b = MOD, x = 1, y = 0;
      while (b > 0) {
        T t = a / b;
        swap(a -= t * b, b);
        swap(x -= t * y, y);
      }
      return ModularInteger(x);
    }
    ModularInteger pow(long long e) const {
      ModularInteger ret = 1, mul = v;
      while (e > 0) {
        if (e & 1) {
          ret *= mul;
        }
        mul *= mul;
        e >>= 1;
      }
      return ret;
    }
    bool operator==(const ModularInteger& p) const { return v == p.v; }
    bool operator!=(const ModularInteger& p) const { return v != p.v; }
    friend ostream& operator<<(ostream& os, const ModularInteger& o) {
      return os << o.v;
    }
    friend istream& operator>>(istream& is, ModularInteger& i) {
      T t;
      cin >> t;
      i = ModularInteger<T, MOD>(t);
      return is;
    }
  };
  using mint = ModularInteger<long long, (long long)1000000007>;
  int N;
  mint arr[15];
  mint dp[1 << 15];
  mint cst[15][15];
  mint mskcst[15][1 << 15];
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> N;
    for (int i = 0; i < N; i++) {
      cin >> arr[i];
    }
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < N; j++) {
        if (i != j) {
          cst[i][j] = arr[i] / (arr[i] + arr[j]);
        }
      }
    }
    for (int i = 0; i < N; i++) {
      for (int m = 0; m < (1 << N); m++) {
        if (m >> i & 1) {
          continue;
        }
        mskcst[i][m] = 1;
        for (int k = 0; k < N; k++) {
          if (m >> k & 1) {
            mskcst[i][m] *= cst[i][k];
          }
        }
      }
    }
    for (int m = 1; m < (1 << N); m++) {
      mint sum = 0;
      for (int t = m; t; t = (m & (t - 1))) {
        mint tot = dp[t];
        for (int k = 0; k < N; k++) {
          if (t >> k & 1) {
            tot *= mskcst[k][m ^ t];
          }
        }
        sum += tot;
      }
      dp[m] = mint(1) - sum;
    }
    mint ans = 0;
    for (int m = 1; m < (1 << N); m++) {
      mint tot = dp[m] * __builtin_popcount(m);
      for (int k = 0; k < N; k++) {
        if (m >> k & 1) {
          tot *= mskcst[k][((1 << N) - 1) ^ m];
        }
      }
      ans += tot;
    }
    cout << ans << "\n";
  }
  