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 maxn = 2e5 + 5;
  const int maxm = 1e6 + 5;
  const int inf = 0x3f3f3f3f;
  const long long mod = 1e9 + 7;
  const double eps = 1e-10;
  int dir[4][2] = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
  mt19937 rnd(
      (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
  long long fast_pow(long long a, long long k) {
    long long miao = 1;
    while (k) {
      if (k & 1) miao *= a, miao %= mod;
      a *= a;
      a %= mod;
      k >>= 1;
    }
    return miao;
  }
  int n;
  long long d[14][14], a[14], f[1 << 14], to[1 << 14][14];
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        d[i][j] = 1LL * a[i] * fast_pow(a[i] + a[j], mod - 2) % mod;
    for (int i = 0; i < (1 << n); i++) {
      for (int j = 0; j < n; j++) {
        to[i][j] = 1;
        for (int k = 0; k < n; k++) {
          if (i & (1 << k)) to[i][j] = 1LL * to[i][j] * d[j][k] % mod;
        }
      }
    }
    long long miao = 0;
    for (int i = 0; i < n; i++) {
      memset(f, 0, sizeof f);
      f[1 << i] = 1;
      for (int j = 0; j < (1 << n); j++) {
        if (j == (1 << i)) continue;
        if (!(j & (1 << i))) continue;
        f[j] = 1;
        for (int u = j; u; u = (u - 1) & j) {
          if (u == j) continue;
          long long cur = f[u];
          for (int s = 0; s < n; s++)
            if ((u ^ j) & (1 << s)) cur = cur * to[u][s] % mod;
          f[j] = (f[j] - cur + mod) % mod;
        }
      }
      miao += f[(1 << n) - 1];
      miao %= mod;
    }
    cout << miao;
    return ~~(0 - 0);
  }
  