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;
  inline int read() {
    int r = 0, f = 1;
    char c = getchar();
    while (c < 48 || c > 57) {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (c >= 48 && c <= 57) r = r * 10 + (c & 15), c = getchar();
    return r * f;
  }
  inline void write(int x) {
    char c[20], len = 0;
    if (!x) {
      putchar('0');
      return;
    }
    if (x < 0) x = -x, putchar('-');
    while (x) c[len++] = x % 10, x /= 10;
    while (len) putchar(c[--len] + 48);
  }
  const int mod = 1e9 + 7;
  struct M {
    int x;
    inline M(int a = 0) : x(a) {}
    inline M operator+(const M& p) const {
      return x + p.x >= mod ? x + p.x - mod : x + p.x;
    }
    inline M operator-() const { return x ? mod - x : 0; }
    inline M operator-(const M& p) const {
      return x - p.x < 0 ? x - p.x + mod : x - p.x;
    }
    inline M operator*(const M& p) const { return (long long)x * p.x % mod; }
    inline bool operator==(const int& p) const { return x == p; }
    inline void operator+=(const M& p) { *this = *this + p; }
    inline void operator-=(const M& p) { *this = *this - p; }
    inline void operator*=(const M& p) { *this = *this * p; }
  };
  inline void write(const M& x) { write(x.x); }
  inline M qpow(M x, int y) {
    M res = 1;
    for (; y; y >>= 1) y & 1 ? (res = res * x) : 1, x *= x;
    return res;
  }
  inline M inv(M x) { return qpow(x, mod - 2); }
  const int N = 14, L = 1 << 14;
  int n, lg[L];
  M a[N], p[N][N], f[N][L], h[N][L];
  inline M G(int S, int T) {
    M res = 1;
    while (S) res *= h[lg[S & -S]][T], S ^= S & -S;
    return res;
  }
  int main() {
    n = read();
    for (int i = 0; i < n; ++i) a[i] = read();
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        if (i ^ j) p[i][j] = a[i] * inv(a[i] + a[j]);
    int all = (1 << n) - 1;
    for (int i = 2; i <= all; ++i) lg[i] = lg[i >> 1] + 1;
    for (int i = 0; i < n; ++i) {
      h[i][0] = 1;
      for (int S = 1; S <= all; ++S)
        if (S >> i & 1 ^ 1) h[i][S] = h[i][S ^ (S & -S)] * p[lg[S & -S]][i];
    }
    for (int i = 0; i < n; ++i)
      for (int S = 1; S <= all; ++S)
        if (S >> i & 1) {
          f[i][S] = 1;
          for (int T = (S - 1) & S; T; T = (T - 1) & S)
            if (T >> i & 1) f[i][S] -= f[i][T] * G(T, S ^ T);
        }
    M ans = 0;
    for (int i = 0; i < n; ++i) ans += f[i][all];
    write(ans), putchar('\n');
  }
  