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 = (long long)1e9 + 7;
  const int SIZE = 1000002;
  const long long INF = 1LL << 30;
  const double eps = 1e-4;
  const double PI = 3.1415926535897932;
  struct ModInt {
    unsigned x;
    ModInt() : x(0) {}
    ModInt(signed sig) : x(sig % MOD) {}
    ModInt(signed long long sig) : x(sig % MOD) {}
    int get() const { return (int)x; }
    ModInt pow(long long p) {
      ModInt res = 1, a = *this;
      while (p) {
        if (p & 1) res *= a;
        a *= a;
        p >>= 1;
      }
      return res;
    }
    ModInt &operator+=(ModInt that) {
      if ((x += that.x) >= MOD) x -= MOD;
      return *this;
    }
    ModInt &operator-=(ModInt that) {
      if ((x += MOD - that.x) >= MOD) x -= MOD;
      return *this;
    }
    ModInt &operator*=(ModInt that) {
      x = (unsigned long long)x * that.x % MOD;
      return *this;
    }
    ModInt &operator/=(ModInt that) { return (*this) *= that.pow(MOD - 2); }
    ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
    ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
    ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
    ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
    bool operator<(ModInt that) const { return x < that.x; }
    friend ostream &operator<<(ostream &os, ModInt a) {
      os << a.x;
      return os;
    }
  };
  ModInt a[15];
  ModInt adj[15][15];
  ModInt p[16384];
  ModInt h[14][16384];
  int n;
  ModInt g(int x, int y) {
    ModInt ans = 1;
    for (int i = 0; i < (n); ++i) {
      if (x >> i & 1) {
        ans *= h[i][y];
      }
    }
    return ans;
  }
  int main() {
    scanf("%d", &(n));
    for (int i = 0; i < (n); ++i) {
      int xx;
      scanf("%d", &xx);
      a[i] = xx;
    }
    for (int i = 0; i < (n); ++i) {
      for (int j = 0; j < (n); ++j) {
        if (i == j) continue;
        adj[i][j] = a[i] / (a[i] + a[j]);
      }
    }
    int bmax = 1 << n;
    for (int i = 0; i < (n); ++i) {
      for (int bm = 0; bm < (bmax); ++bm) {
        h[i][bm] = 1;
        for (int bi = 0; bi < (n); ++bi) {
          if (bm >> bi & 1) {
            h[i][bm] *= adj[i][bi];
          }
        }
      }
    }
    p[0] = 1;
    for (int bm = (1); bm < (bmax); ++bm) {
      p[bm] = 1;
      for (int bm2 = bm; bm2; bm2 = (bm2 - 1) & bm) {
        if (bm2 == bm) continue;
        p[bm] -= p[bm2] * g(bm2, bm ^ bm2);
        if (bm2 == 0) break;
      }
    }
    ModInt ans = 0;
    for (int bm = (1); bm < (bmax); ++bm) {
      ModInt f = p[bm] * g(bm, (bmax - 1) ^ bm);
      ans += f * __builtin_popcountll(bm);
    }
    printf("%u", ans.x);
  }
  