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 int mod = 1e9 + 7;
  long long inv(long long a) {
    long long u = 0, v = 1, b = mod;
    while (a) {
      long long t = b / a;
      b -= t * a, u -= t * v;
      swap(a, b);
      swap(u, v);
    }
    u %= mod;
    if (u < 0) u += mod;
    return u;
  }
  struct mint {
    int x;
    mint() : x(0) {}
    mint(int x) : x(x) {}
    mint(long long x) : x() {
      if (x >= mod || x < -mod) x %= mod;
      if (x < 0) x += mod;
      this->x = 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 = (long long)x * rhs.x % mod;
      return *this;
    }
    mint &operator/=(const mint &rhs) {
      x = (long long)x * inv(rhs.x) % mod;
      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) {
      mint ret(lhs);
      return ret += rhs;
    }
    friend mint operator-(const mint &lhs, const mint &rhs) {
      mint ret(lhs);
      return ret -= rhs;
    }
    friend mint operator*(const mint &lhs, const mint &rhs) {
      mint ret(lhs);
      return ret *= rhs;
    }
    friend mint operator/(const mint &lhs, const mint &rhs) {
      mint ret(lhs);
      return ret /= rhs;
    }
    friend ostream &operator<<(ostream &os, const mint &rhs) {
      os << rhs.x;
      return os;
    }
    friend istream &operator>>(istream &is, mint &rhs) {
      long long a;
      is >> a;
      rhs = a;
      return is;
    }
  };
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    mint a[n];
    for (int i = 0; i < n; ++i) cin >> a[i];
    mint dp[1 << n], mul[1 << n][n], inner[1 << n][2];
    for (int i = 0; i < 1 << n; ++i) {
      mul[i][0] = mul[i][1] = 1;
      for (int j = 0; j < n; ++j) {
        if (i >> j & 1) mul[i][1] = mul[i][1] * a[j];
      }
      for (int j = 2; j < n; ++j) {
        mul[i][j] = mul[i][j - 1] * mul[i][1];
      }
      if (i > 0) {
        int b = 31 - __builtin_clz(i);
        inner[i][0] = inner[i ^ (1 << b)][0];
        for (int j = 0; j < b; ++j) {
          if (i >> j & 1) inner[i][0] = inner[i][0] * (a[b] + a[j]);
        }
      } else {
        inner[i][0] = 1;
      }
      inner[i][1] = 1 / inner[i][0];
    }
    for (int i = 1; i < 1 << n; ++i) {
      dp[i] = 1;
      for (int j = (i - 1) & i; j; j = (j - 1) & i) {
        dp[i] -= dp[j] * mul[j][__builtin_popcount(i ^ j)] * inner[j][0] *
                 inner[i ^ j][0] * inner[i][1];
      }
    }
    mint ans = 0;
    for (int i = 1; i < 1 << n; ++i) {
      int p = (1 << n) - 1;
      ans += mul[i][__builtin_popcount(i ^ p)] * inner[i][0] * inner[i ^ p][0] *
             inner[p][1] * dp[i] * __builtin_popcount(i);
    }
    cout << ans << '\n';
  }
  