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;
  template <long long mod>
  struct mint {
    long long x;
    static long long norm(long long x) {
      if (x >= mod + mod || x < -mod) x %= mod;
      if (x > mod) x -= mod;
      if (x < 0) x += mod;
      return x;
    }
    static long long inv(long long a) {
      long long u = 0, v = 1, m = mod;
      while (a != 0) {
        long long t = m / a;
        m -= t * a;
        swap(a, m);
        u -= t * v;
        swap(u, v);
      }
      return norm(u);
    }
    mint() : x(0) {}
    template <typename T>
    mint(T x) : x(norm(x)) {}
    long long operator()() { return x; }
    mint inv() { return mint(inv(x)); }
    mint &operator-=(const mint &rhs) {
      x -= rhs.x;
      if (x < 0) x += mod;
      return *this;
    }
    mint &operator+=(const mint &rhs) {
      x += rhs.x;
      if (x >= mod) x -= mod;
      return *this;
    }
    mint &operator*=(const mint &rhs) {
      x = norm(x * rhs.x);
      return *this;
    }
    mint &operator/=(const mint &rhs) {
      x = norm(x * inv(rhs.x));
      return *this;
    }
    mint &operator++() { return *this += 1; }
    mint &operator--() { return *this -= 1; }
    mint &operator++(int) {
      x += 1;
      return *this;
    }
    mint &operator--(int) {
      x -= 1;
      return *this;
    }
    bool operator<(const mint &rhs) { return x < rhs.x; }
    bool operator==(const mint &rhs) { return x == rhs.x; }
    bool operator!=(const mint &rhs) { return x != rhs.x; }
    friend mint operator/(const mint &lhs, const mint &rhs) {
      return mint(lhs.x * inv(rhs.x));
    }
    friend mint operator*(const mint &lhs, const mint &rhs) {
      return mint(lhs.x * rhs.x);
    }
    friend mint operator+(const mint &lhs, const mint &rhs) {
      return mint(lhs.x + rhs.x);
    }
    friend mint operator-(const mint &lhs, const mint &rhs) {
      return mint(lhs.x - rhs.x);
    }
    friend ostream &operator<<(ostream &os, const mint &rhs) {
      os << rhs.x;
      return os;
    }
    friend istream &operator>>(istream &is, mint &rhs) {
      is >> rhs.x;
      rhs.x = norm(rhs.x);
      return is;
    }
  };
  const int mod = 1e9 + 7;
  using Z = mint<mod>;
  Z fpow(Z a, long long b) {
    Z r = 1;
    for (; b; b >>= 1) {
      if (b & 1) r = r * a;
      a = a * a;
    }
    return r;
  }
  Z comb(int n, int k) {
    if (n < k || k < 0) return 0;
    static vector<Z> f(1, 1), rf(1, 1);
    int p = f.size();
    if (p < n) f.resize(n + 1), rf.resize(n + 1);
    for (int i = p; i <= n; ++i) f[i] = f[i - 1] * i;
    rf[n] = 1 / f[n];
    for (int i = n - 1; i >= p; --i) rf[i] = rf[i + 1] * (i + 1);
    return f[n] * rf[k] * rf[n - k];
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    vector<Z> a(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    vector<vector<Z>> win(n, vector<Z>(n));
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (i == j) continue;
        win[i][j] = a[i] * fpow(a[i] + a[j], mod - 2);
      }
    }
    vector<Z> dp(1 << n);
    for (int i = 1; i < 1 << n; ++i) {
      dp[i] = 1;
      for (int j = (i - 1) & i; j; j = (j - 1) & i) {
        Z tmp = 1;
        for (int k = 0; k < n; ++k) {
          if ((i >> k & 1) && (j >> k & 1))
            for (int p = 0; p < n; ++p) {
              if ((i >> p & 1) && !(j >> p & 1)) {
                tmp = tmp * win[k][p];
              }
            }
        }
        dp[i] -= dp[j] * tmp;
      }
    }
    Z ans = 0;
    for (int i = 1; i < 1 << n; ++i) {
      Z tmp = 1;
      for (int j = 0; j < n; ++j) {
        if (i >> j & 1)
          for (int k = 0; k < n; ++k) {
            if (!(i >> k & 1)) tmp = tmp * win[j][k];
          }
      }
      ans += tmp * dp[i] * __builtin_popcount(i);
    }
    cout << ans << '\n';
  }
  