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;
  long long mmod(long long x, long long y) {
    if (x >= 0LL)
      return x % y;
    else
      return (y - (-x) % y) % y;
  }
  long long ddiv(long long x, long long y) {
    x -= mmod(x, y);
    return x / y;
  }
  long long binpow(long long a, long long b, long long mod) {
    if (a == 0) {
      return 0;
    }
    long long pows[65];
    pows[0] = a;
    for (int i = 1; i < 65; ++i) {
      pows[i] = mmod(pows[i - 1] * pows[i - 1], mod);
    }
    long long ans = 1;
    for (int i = 0; b != 0; b >>= 1, ++i) {
      if (b % 2 == 1) {
        ans = mmod(ans * pows[i], mod);
      }
    }
    return ans;
  }
  long long GCD(long long a, long long b) {
    while (a != 0) {
      b = b % a;
      swap(a, b);
    }
    return b;
  }
  const int M = 1000000007;
  int bmRange;
  vector<int> a;
  struct frac {
    long long num;
    long long denom;
    frac() : num(0), denom(1) {}
    frac(long long n, long long d) : num(n), denom(d) {}
    frac operator*(const frac& other) const {
      return {mmod(num * other.num, M), mmod(denom * other.denom, M)};
    }
    frac operator+(const frac& other) const {
      return {mmod(num * other.denom + denom * other.num, M),
              mmod(denom * other.denom, M)};
    }
    frac operator-() const { return {mmod(denom - num, M), denom}; }
    long long get() const { return mmod(num * binpow(denom, M - 2, M), M); }
  };
  vector<frac> probs;
  frac get_prob(int submask, int bitmask) {
    frac res(1, 1);
    for (int i = 0; i < 14; ++i)
      for (int j = 0; j < 14; ++j) {
        if ((submask >> i) & 1 && (bitmask >> j) & 1 && !((submask >> j) & 1)) {
          res = res * frac(a[i], a[i] + a[j]);
        }
      }
    return res;
  }
  frac res{0, 1};
  void apply(int bitmask) {
    for (int submask = bitmask & (bitmask - 1); submask != 0;
         submask = bitmask & (submask - 1)) {
      probs[bitmask] =
          probs[bitmask] + (-probs[submask]) * get_prob(submask, bitmask);
    }
    int memc = 0;
    for (int i = 0; i < 14; ++i) {
      if ((bitmask >> i) & 1) ++memc;
    }
    res =
        res + (-probs[bitmask]) * get_prob(bitmask, bmRange - 1) * frac(memc, 1);
  }
  void solve() {
    int n;
    cin >> n;
    bmRange = (1 << n);
    a.resize(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    probs.resize(bmRange, {0, 1});
    for (int i = 0; i < bmRange; ++i) apply(i);
    cout << res.get() << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int t = 1;
    for (int k = 0; k < t; ++k) solve();
    return 0;
  }
  