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 P = 1e9 + 7, N = 17, M = 16389, O = 131;
  int qp(int a, int b) {
    int r = 1;
    for (; b; b >>= 1, a = a * 1ll * a % P)
      if (b & 1) r = r * 1ll * a % P;
    return r;
  }
  int a[N], e[N][N], p[M], n, ll[O][O], lr[O][O], rl[O][O], rr[O][O];
  int g(int x, int y) {
    int i, j, s = 1;
    for (i = 0; i < n; ++i)
      if (x >> i & 1)
        for (j = 0; j < n; ++j)
          if (y >> j & 1) s = s * 1ll * e[i][j] % P;
    return s;
  }
  int h(int x, int y) {
    return ll[x & 127][y & 127] * 1ll * lr[x & 127][y >> 7] % P *
           rl[x >> 7][y & 127] % P * rr[x >> 7][y >> 7] % P;
  }
  int main() {
    int m, i, j, w = 0;
    scanf("%d", &n);
    for (i = 0; i < n; ++i) scanf("%d", a + i);
    for (i = 0; i < n; ++i)
      for (j = 0; j < n; ++j) e[i][j] = a[i] * 1ll * qp(a[i] + a[j], P - 2) % P;
    for (i = 0; i < O; ++i)
      for (j = 0; j < O; ++j) ll[i][j] = lr[i][j] = rl[i][j] = rr[i][j] = 1;
    for (i = 0, m = 1 << min(n, 7); i < m; ++i)
      for (j = 0; j < m; ++j)
        if (!(i & j)) ll[i][j] = g(i, j);
    if (n > 7) {
      for (i = 0, m = 1 << n - 7; i < 128; ++i)
        for (j = 0; j < m; ++j) lr[i][j] = g(i, j << 7);
      for (i = 0; i < m; ++i)
        for (j = 0; j < 128; ++j) rl[i][j] = g(i << 7, j);
      for (i = 0; i < m; ++i)
        for (j = 0; j < m; ++j)
          if (!(i & j)) rr[i][j] = g(i << 7, j << 7);
    }
    for (i = 1, m = 1 << n; i < m;
         w = (w + p[i] * 1ll * h(i, m - 1 ^ i) % P * __builtin_popcount(i)) % P,
        ++i)
      for (p[i] = 1, j = i & i - 1; j; j = j - 1 & i)
        p[i] = (p[i] - p[j] * 1ll * h(j, i ^ j)) % P;
    printf("%d", (w + P) % P);
    return 0;
  }
  