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;
  struct mi {
    typedef decay<decltype(MOD)>::type T;
    T v;
    explicit operator T() const { return v; }
    mi() { v = 0; }
    mi(long long _v) {
      v = (-MOD < _v && _v < MOD) ? _v : _v % MOD;
      if (v < 0) v += MOD;
    }
    friend bool operator==(const mi& a, const mi& b) { return a.v == b.v; }
    friend bool operator!=(const mi& a, const mi& b) { return !(a == b); }
    friend bool operator<(const mi& a, const mi& b) { return a.v < b.v; }
    mi& operator+=(const mi& m) {
      if ((v += m.v) >= MOD) v -= MOD;
      return *this;
    }
    mi& operator-=(const mi& m) {
      if ((v -= m.v) < 0) v += MOD;
      return *this;
    }
    mi& operator*=(const mi& m) {
      v = (long long)v * m.v % MOD;
      return *this;
    }
    mi& operator/=(const mi& m) { return (*this) *= inv(m); }
    friend mi pow(mi a, long long p) {
      mi ans = 1;
      assert(p >= 0);
      for (; p; p /= 2, a *= a)
        if (p & 1) ans *= a;
      return ans;
    }
    friend mi inv(const mi& a) {
      assert(a.v != 0);
      return pow(a, MOD - 2);
    }
    mi operator-() const { return mi(-v); }
    mi& operator++() { return *this += 1; }
    mi& operator--() { return *this -= 1; }
    friend mi operator+(mi a, const mi& b) { return a += b; }
    friend mi operator-(mi a, const mi& b) { return a -= b; }
    friend mi operator*(mi a, const mi& b) { return a *= b; }
    friend mi operator/(mi a, const mi& b) { return a /= b; }
  };
  const int N = 15, M = (1 << 14) + 5;
  map<vector<int>, int> mp;
  void pre(vector<int> vs) {}
  int n, a[N];
  int bcnt[M];
  vector<int> bt[M];
  mi win[N][N];
  mi dp[M];
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n;
    for (int i = 0; i < (n); ++i) cin >> a[i];
    for (int i = 0; i < (n); ++i) {
      for (int j = 0; j < (n); ++j) {
        win[i][j] = mi(a[i]) / (mi(a[i]) + mi(a[j]));
      }
    }
    for (int i = 1; i < (1 << n); ++i) bcnt[i] = bcnt[i / 2] + (i & 1);
    for (int mask = 0; mask < (1 << n); ++mask) {
      for (int b = 0; b < (n); ++b)
        if (mask >> b & 1) bt[mask].push_back(b);
    }
    mi fres = 0;
    for (int mask = 1; mask < (1 << n); ++mask) {
      dp[mask] = 1;
      for (int sub = mask;; sub = (sub - 1) & mask) {
        if (sub == mask) continue;
        int rem = mask ^ sub;
        mi s = dp[sub];
        for (int b : bt[sub]) {
          for (int r : bt[rem]) {
            s *= win[b][r];
          }
        }
        dp[mask] -= s;
        if (sub == 0) break;
      }
      mi s = dp[mask];
      int rem = ((1 << n) - 1) ^ mask;
      for (int b : bt[mask]) {
        for (int r : bt[rem]) {
          s *= win[b][r];
        }
      }
      fres += s * bcnt[mask];
    }
    cout << int(fres) << '\n';
  }
  