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 char gc() {
    static char buf[100000], *p1 = buf, *p2 = buf;
    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
               ? EOF
               : *p1++;
  }
  inline long long read() {
    char c = getchar();
    long long su = 0, f = 1;
    for (; c < '0' || c > '9'; c = getchar())
      if (c == '-') f = -1;
    for (; c >= '0' && c <= '9'; c = getchar()) su = su * 10 + c - '0';
    return su * f;
  }
  inline void write(long long x) {
    if (x < 0) {
      putchar('-');
      write(-x);
      return;
    }
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
  }
  inline void writesp(long long x) { write(x), putchar(' '); }
  inline void writeln(long long x) {
    write(x);
    putchar('\n');
  }
  long long P[15][15], f[1 << 15], ans, n, a[15];
  long long quickPower(long long x, long long y) {
    long long res = 1;
    while (y) {
      if (y % 2) res = res * x % 1000000007;
      x = x * x % 1000000007;
      y >>= 1;
    }
    return res;
  }
  long long work(long long x, long long y) {
    vector<long long> g;
    long long res = 1;
    for (long long i = 1; i <= n; i++) {
      if ((y >> (i - 1)) % 2) g.push_back(i);
    }
    for (long long i = 1; i <= n; i++)
      if ((x >> (i - 1)) % 2) {
        long long now = 1;
        for (auto u : g) {
          now = now * P[u][i] % 1000000007;
        }
        res = res * now % 1000000007;
      }
    return res;
  }
  signed main() {
    n = read();
    for (long long i = 1; i <= n; i++) a[i] = read();
    for (long long i = 1; i <= n; i++)
      for (long long j = 1; j <= n; j++)
        P[i][j] = a[i] * quickPower(a[i] + a[j], 1000000007 - 2) % 1000000007;
    for (long long stat = 1; stat < (1 << n); stat++) {
      f[stat] = 1;
      for (long long S = (stat - 1) & stat; S; S = (S - 1) & stat) {
        f[stat] = (f[stat] - f[S] * work(stat ^ S, S) % 1000000007 + 1000000007) %
                  1000000007;
        if (S == 0) break;
      }
      ans = (ans + __builtin_popcount(stat) * f[stat] % 1000000007 *
                       work(((1 << n) - 1) ^ stat, stat)) %
            1000000007;
    }
    writeln(ans);
  }
  