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 N = 15;
  const int maxn = 2e5 + 10;
  const int mod = 1e9 + 7;
  inline int read() {
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      x = (x << 1) + (x << 3) + c - '0';
      c = getchar();
    }
    return x * f;
  }
  int n, a[N], all, ans, f[1 << N], g[1 << N], iv[N][N], cnt[1 << N], lg[1 << N],
      lw[1 << N];
  inline int ksm(int x, int y) {
    int res = 1;
    while (y) {
      if (y & 1) res = 1ll * res * x % mod;
      x = 1ll * x * x % mod;
      y >>= 1;
    }
    return res;
  }
  inline int lwb(int x) { return x & (-x); }
  inline void add(int &x, int y) { x = (x + y >= mod ? x + y - mod : x + y); }
  inline void sub(int &x, int y) { x = (x < y ? x - y + mod : x - y); }
  int main() {
    n = read();
    all = (1 << n) - 1;
    for (int i = 0; i < n; i++) a[i] = read();
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) iv[i][j] = ksm(a[i] + a[j], mod - 2);
    for (int i = 1; i <= all; i++) {
      g[i] = 1;
      for (int j = 0; j < n; j++)
        if (i >> j & 1)
          for (int k = 0; k < n; k++)
            if (~i >> k & 1) g[i] = 1ll * g[i] * a[j] % mod * iv[j][k] % mod;
    }
    lg[0] = -1;
    for (int i = 1; i <= all; i++) {
      cnt[i] = cnt[i >> 1] + (i & 1);
      lg[i] = lg[i >> 1] + 1;
      lw[i] = lwb(i);
      lw[i] = lg[lw[i]];
    }
    for (int i = 1; i <= all; i++) {
      f[i] = g[i];
      for (int s = ((i - 1) & i); s; s = ((s - 1) & i)) {
        int cur = 1;
        for (int j = s; j; j &= (j - 1))
          for (int k = (i ^ s); k; k &= (k - 1))
            cur = 1ll * cur * a[lw[k]] % mod * iv[lw[j]][lw[k]] % mod;
        sub(f[i], 1ll * f[s] * g[i ^ s] % mod * ksm(cur, mod - 2) % mod);
      }
      add(ans, 1ll * cnt[i] * f[i] % mod);
    }
    printf("%d\n", ans);
    return 0;
  }
  