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 MOD = 1e9 + 7;
  int n;
  long long a[20], dp[1 << 20], dp2[1 << 20], vs[20][20], ans = 0;
  long long pw(long long t, long long e) {
    if (!e) return 1;
    long long ret = pw(t, e / 2);
    if (e & 1) return ret * ret % MOD * t % MOD;
    return ret * ret % MOD;
  }
  long long INV(long long x) { return pw(x, MOD - 2); }
  void solve(int x) {
    long long val = 1;
    for (int z = 1; z < (1 << n) - 1; z++)
      if (!(z & (1 << x))) {
        val -= dp2[z];
        if (val < 0) val += MOD;
      }
    ans += val;
    if (ans >= MOD) ans -= MOD;
  }
  void init() {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        vs[i][j] = a[i] * INV(a[i] + a[j]) % MOD;
      }
    }
    for (int z = 0; z < (1 << n); z++) {
      dp[z] = 1;
      for (int i = 0; i < n; i++)
        if (z & (1 << i)) {
          for (int j = 0; j < n; j++)
            if (!(z & (1 << j))) {
              dp[z] *= vs[i][j];
              dp[z] %= MOD;
            }
        }
    }
    for (int z = 1; z < (1 << n) - 1; z++) {
      dp2[z] = dp[z];
      for (int i = z; i > 0; i = (i - 1) & z)
        if (i != z) {
          long long tmp = dp2[i];
          for (int j = 0; j < n; j++)
            if (z & (1 << j) && !(i & (1 << j))) {
              for (int k = 0; k < n; k++)
                if (!(z & (1 << k))) {
                  tmp *= vs[j][k];
                  tmp %= MOD;
                }
            }
          dp2[z] -= tmp;
          if (dp2[z] < 0) dp2[z] += MOD;
        }
    }
  }
  int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) scanf("%lld", a + i);
    if (n == 1) {
      printf("1\n");
      return 0;
    }
    init();
    for (int i = 0; i < n; i++) solve(i);
    printf("%lld\n", ans);
    return 0;
  }
  