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;
  void per() { cerr << endl; }
  template <typename Head, typename... Tail>
  void per(Head H, Tail... T) {
    cerr << H << ' ';
    per(T...);
  }
  template <class T>
  bool uin(T& a, T b) {
    return a > b ? (a = b, true) : false;
  }
  template <class T>
  bool uax(T& a, T b) {
    return a < b ? (a = b, true) : false;
  }
  template <class U, class V>
  ostream& operator<<(ostream& out, const pair<U, V>& a) {
    return out << "(" << a.first << ", " << a.second << ")";
  }
  template <class U, class V>
  istream& operator>>(istream& in, pair<U, V>& a) {
    return in >> a.first >> a.second;
  }
  template <typename W,
            typename T = typename enable_if<!is_same<W, string>::value,
                                            typename W::value_type>::type>
  ostream& operator<<(ostream& out, const W& v) {
    out << "{ ";
    for (const auto& first : v) out << first << ", ";
    return out << '}';
  }
  template <class T>
  void readArr(T from, T to) {
    for (auto i = from; i != to; ++i) cin >> *i;
  }
  mt19937 mrand(1337);
  unsigned int myRand32() { return mrand() & (unsigned int)(-1); }
  unsigned long long myRand64() {
    return ((unsigned long long)myRand32() << 32) ^ myRand32();
  }
  const int mod = 1000000007;
  void add(int& a, int b) {
    a += b;
    if (a >= mod) a -= mod;
  }
  void dec(int& a, int b) {
    a -= b;
    if (a < 0) a += mod;
  }
  int mult(int a, int b) { return a * (long long)b % mod; }
  int bp(int a, int b) {
    int res = 1;
    while (b > 0) {
      if (b & 1) res = mult(res, a);
      a = mult(a, a);
      b >>= 1;
    }
    return res;
  }
  const int N = 14;
  int dp[1 << N];
  int pp[N][N];
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    vector<int> v(n);
    for (auto& first : v) cin >> first;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        pp[i][j] = mult(v[i], bp(v[i] + v[j], mod - 2));
      }
    }
    int ans = 0;
    for (int mask = 1; mask < (1 << n); ++mask) {
      int p = 1;
      vector<int> a, b;
      for (int i = 0; i < n; ++i) {
        if ((1 << i) & mask)
          a.push_back(i);
        else
          b.push_back(i);
      }
      for (auto first : a)
        for (auto second : b) p = mult(p, pp[first][second]);
      for (int s = mask; s; s = (s - 1) & mask) {
        if (s == mask) continue;
        int t = dp[mask ^ s];
        vector<int> c;
        for (int i = 0; i < n; ++i) {
          if ((1 << i) & s) c.push_back(i);
        }
        for (auto first : c)
          for (auto second : b) t = mult(t, pp[first][second]);
        dec(p, t);
      }
      dp[mask] = p;
      add(ans, mult(p, __builtin_popcount(mask)));
    }
    cout << ans;
  }
  