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 x = 0, f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '0') f = -1;
      ch = getchar();
    }
    while (isdigit(ch)) {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    return x * f;
  }
  const int MAXN = 5000006, Mo = 1000000007;
  void add(int &x, int y) {
    x += y;
    if (x >= Mo) x -= Mo;
  }
  void sub(int &x, int y) {
    x -= y;
    if (x < 0) x += Mo;
  }
  int inv(int x) {
    int ret = 1;
    for (int y = Mo - 2; y; y >>= 1, x = 1ll * x * x % Mo)
      if (y & 1) ret = 1ll * ret * x % Mo;
    return ret;
  }
  int n, m, a[14], b[14][14], g[14][1 << 14], g_inv[14][1 << 14];
  int idx[1 << 14], idy[1 << 14], f[MAXN], f_inv[MAXN], p[1 << 14];
  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++) b[i][j] = 1ll * a[i] * inv(a[i] + a[j]) % Mo;
    for (int i = 0; i < (n); i++)
      for (int j = 0; j < ((1 << n)); j++) {
        if (j == 0)
          g[i][j] = g_inv[i][j] = 1;
        else
          g[i][j] = 1ll * g[i][j & (j - 1)] * b[i][__builtin_ctz(j)] % Mo,
          g_inv[i][j] = inv(g[i][j]);
      }
    for (int k = 0; k < ((1 << n)); k++) {
      idx[k] = idy[k] = 0;
      for (int i = 0, j = 1; i < n; ++i, j *= 3)
        if ((k >> i) & 1) idx[k] += j * 2, idy[k] += j;
    }
    for (int i = 0; i < ((1 << n)); i++) {
      for (int j = i; j; j = ((j - 1) & i)) {
        int k = (i ^ j);
        int now = idx[k] + idy[j], tmp = idx[k & (k - 1)] + idy[j];
        if (k == 0) {
          f[now] = f_inv[now] = 1;
          continue;
        }
        int __k = __builtin_ctz(k);
        f[now] = 1ll * f[tmp] * g[__k][j] % Mo;
        f_inv[now] = 1ll * f_inv[tmp] * g_inv[__k][j] % Mo;
      }
      f[idx[i]] = f_inv[idx[i]] = 1;
    }
    int ans = 0, all = (1 << n) - 1;
    for (int win = 0; win < ((1 << n)); win++) {
      p[win] = 1;
      for (int s = ((win - 1) & win); s; s = ((s - 1) & win))
        sub(p[win], 1ll * p[s] * f_inv[idx[s] + idy[all - win]] % Mo);
      p[win] = 1ll * p[win] * f[idx[win] + idy[all - win]] % Mo;
      add(ans, 1ll * p[win] * __builtin_popcount(win) % Mo);
    }
    printf("%d\n", ans);
    return 0;
  }
  