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;
  namespace IO {
  char in[1 << 24];
  char const *o;
  void init_in() {
    o = in;
    in[fread(in, 1, sizeof(in) - 4, stdin)] = 0;
  }
  int rd() {
    unsigned u = 0, s = 0;
    while (*o && *o <= 32) ++o;
    if (*o == '-')
      s = ~s, ++o;
    else if (*o == '+')
      ++o;
    while (*o >= '0' && *o <= '9') u = (u << 3) + (u << 1) + (*o++ - '0');
    return static_cast<int>((u ^ s) + !!s);
  }
  char *rdstr(char *s) {
    while (*o && *o <= 32) ++o;
    while (*o > 32) *s++ = *o++;
    return s;
  }
  }  // namespace IO
  template <typename T>
  void chkmax(T &a, const T &b) {
    a = max(a, b);
  }
  template <typename T>
  void chkmin(T &a, const T &b) {
    a = min(a, b);
  }
  pair<int, int> operator+(const pair<int, int> &a, const pair<int, int> &b) {
    return make_pair(a.first + b.first, a.second + b.second);
  }
  using IO::rd;
  const int N = 14, S = 1 << N;
  int a[N];
  int n;
  const long long mod = 1e9 + 7;
  long long b[N][N];
  long long h[N][S], pr[S];
  inline void reduce(long long &x) {
    if (x >= mod) x -= mod;
  }
  inline long long g(int s, int t) {
    long long ret = 1;
    for (int i = 0; i < n; ++i) {
      if (s & (1 << i)) ret = ret * h[i][t] % mod;
    }
    return ret;
  }
  int Q_pow(long long x, int y) {
    long long ret = 1;
    while (y) {
      if (y & 1) ret = ret * x % mod;
      x = x * x % mod;
      y >>= 1;
    }
    return ret;
  }
  inline int inv(int x) { return Q_pow(x, mod - 2); }
  void solve() {
    n = rd();
    int m = 1 << n;
    for (int i = 0; i < n; ++i) a[i] = rd();
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < i; ++j) {
        long long x = inv(a[i] + a[j]);
        b[i][j] = a[i] * x % mod;
        b[j][i] = a[j] * x % mod;
      }
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j) {
        h[i][j] = 1;
        for (int k = 0; k < n; ++k) {
          if (k != i && j & (1 << k)) h[i][j] = h[i][j] * b[i][k] % mod;
        }
      }
    int mask = m - 1;
    for (int i = 1; i < m; ++i) {
      pr[i] = 1;
      for (int j = (i - 1) & i; j; j = (j - 1) & i) {
        reduce(pr[i] += mod - (g(j, i ^ j) * pr[j] % mod));
      }
    }
    long long ans = 0;
    for (int i = 1; i < m; ++i) {
      ans += pr[i] * g(i, mask ^ i) % mod * __builtin_popcount(i) % mod;
    }
    cout << ans % mod << '\n';
  }
  int main() {
    IO::init_in();
    solve();
    return 0;
  }
  