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 long long mod = static_cast<long long>(1e9 + 7);
  long long P, Q = 1;
  int n;
  long long a[15];
  pair<long long, long long> f[15][1 << 14];
  pair<long long, long long> g[5000000];
  pair<long long, long long> p[1 << 14];
  pair<long long, long long> h[1 << 14];
  int count(int x) {
    if (!x) return 0;
    return (x & 1) + count(x >> 1);
  }
  void Init_F() {
    for (int i = 1; i <= n; ++i) {
      for (int s = 0; s < (1 << n); ++s) {
        if ((1 << (i - 1)) & s) {
          continue;
        }
        f[i][s] = make_pair(1, 1);
        for (int j = 1; j <= n; ++j) {
          if ((1 << (j - 1)) & s) {
            f[i][s].first = f[i][s].first * a[i] % mod;
            f[i][s].second = f[i][s].second * (a[i] + a[j]) % mod;
          }
        }
      }
    }
  }
  int GetIndexBase3(int left, int right) {
    int idx = 0;
    for (int i = n; i > 0; --i) {
      idx *= 3;
      if (left & (1 << (i - 1))) {
        idx += 1;
      } else if (right & (1 << (i - 1))) {
        idx += 2;
      }
    }
    return idx;
  }
  void Recursively_Init_G(int left, int right, int cur) {
    if (cur > n) {
      int idx = GetIndexBase3(left, right);
      g[idx] = make_pair(1, 1);
      for (int i = 1; i <= n; ++i) {
        if ((1 << (i - 1)) & left) {
          g[idx].first = g[idx].first * f[i][right].first % mod;
          g[idx].second = g[idx].second * f[i][right].second % mod;
        }
      }
    } else {
      Recursively_Init_G(left, right, cur + 1);
      Recursively_Init_G(left + (1 << (cur - 1)), right, cur + 1);
      Recursively_Init_G(left, right + (1 << (cur - 1)), cur + 1);
    }
  }
  void Init_P() {
    for (int i = 1; i < (1 << n); ++i) {
      p[i] = make_pair(1, 1);
      for (int j = 1; j < i; ++j) {
        if ((i | j) == i) {
          long long upper = p[j].first * g[GetIndexBase3(j, i - j)].first % mod;
          long long lower = p[j].second * g[GetIndexBase3(j, i - j)].second % mod;
          p[i].first =
              ((p[i].first * lower % mod - upper * p[i].second % mod) % mod +
               mod) %
              mod;
          p[i].second = p[i].second * lower % mod;
        }
      }
    }
  }
  void Solve() {
    Init_F();
    Recursively_Init_G(0, 0, 1);
    Init_P();
    for (int i = 1; i < (1 << n); ++i) {
      int j = (1 << n) - 1 - i;
      long long upper =
          count(i) * p[i].first % mod * g[GetIndexBase3(i, j)].first % mod;
      long long lower = p[i].second * g[GetIndexBase3(i, j)].second % mod;
      P = (P * lower + Q * upper) % mod;
      Q = Q * lower % mod;
    }
  }
  long long Pow(long long x, long long y) {
    long long res = 1;
    while (y) {
      if (y & 1) {
        res = res * x % mod;
      }
      x = x * x % mod;
      y >>= 1;
    }
    return res;
  }
  int main() {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    Solve();
    cout << P * Pow(Q, mod - 2) % mod << endl;
    return 0;
  }
  