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;
  template <typename T>
  void chkMax(T &x, T y) {
    if (y > x) x = y;
  }
  template <typename T>
  void chkMin(T &x, T y) {
    if (y < x) x = y;
  }
  template <typename T>
  void inline read(T &x) {
    int f = 1;
    x = 0;
    char s = getchar();
    while (s < '0' || s > '9') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
  }
  const int N = 14, P = 1e9 + 7;
  int inline power(int a, int b) {
    int res = 1;
    while (b) {
      if (b & 1) res = (long long)res * a % P;
      a = (long long)a * a % P;
      b >>= 1;
    }
    return res;
  }
  void inline add(int &x, int y) {
    x += y;
    if (x >= P) x -= P;
  }
  int n, a[N], w[N][N], wt[N][1 << N], s, ans, f[1 << N];
  int inline ban(int x, int y) {
    int res = 1;
    for (int i = 0; i < n; i++) {
      if (y >> i & 1) res = 1ll * res * wt[i][x] % P;
    }
    return res;
  }
  void inline work() {
    memset(f, 0, sizeof f);
    f[1 << s] = 1;
    for (int i = 0; i < (1 << n); i++) {
      if (!(i >> s & 1)) continue;
      if (i == (1 << s)) continue;
      f[i] = 1;
      for (int j = i - 1; j; j = (j - 1) & i) {
        if (j >> s & 1) continue;
        f[i] = (f[i] - 1ll * ban(i ^ j, j) * f[i ^ j] % P + P) % P;
      }
    }
    add(ans, f[(1 << n) - 1]);
  }
  int main() {
    read(n);
    for (int i = 0; i < n; i++) read(a[i]);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (i == j) continue;
        int inv = power(a[i] + a[j], P - 2);
        w[i][j] = 1ll * a[i] * inv % P;
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < (1 << n); j++) {
        wt[i][j] = 1;
        for (int k = 0; k < n; k++)
          if (j >> k & 1) wt[i][j] = 1ll * wt[i][j] * w[i][k] % P;
      }
    }
    for (s = 0; s < n; s++) work();
    printf("%d\n", ans);
    return 0;
  }
  