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;
  long long ksm(long long x, long long y) {
    long long ret = 1;
    for (; y; y >>= 1, x = x * x % 1000000007)
      if (y & 1) ret = ret * x % 1000000007;
    return ret;
  }
  long long dp[1 << 15], ans, F[1 << 15], a[15], ny[15][15], sz[1 << 15],
      f[1 << 15], n, i, j;
  long long cal(int st1, int st2) {
    long long ret = 1;
    int i, j;
    for (i = 0; i < n; ++i) {
      if ((st1 >> i) & 1) {
        for (j = 0; j < n; ++j) {
          if ((st2 >> j) & 1) {
            (ret *= ny[i][j]) %= 1000000007;
          }
        }
      }
    }
    cout << st1 << " " << st2 << " " << ret << ' ' << F[st1] << " " << F[st2]
         << "\n";
    return ret * (ksm(F[st1], sz[st2]) % 1000000007 + ksm(F[st2], sz[st1])) %
           1000000007;
  }
  long long CAL(int st1, int st2) {
    long long ret = 1;
    int i, j;
    for (i = 0; i < n; ++i) {
      if ((st1 >> i) & 1) {
        for (j = 0; j < n; ++j) {
          if ((st2 >> j) & 1) {
            (ret *= ny[i][j]) %= 1000000007;
          }
        }
      }
    }
    return ret * ksm(F[st1], sz[st2]) % 1000000007;
  }
  int mx, x, f1;
  signed main() {
    cin >> n;
    for (i = 0; i < n; ++i) cin >> a[i];
    for (i = 0; i < n; ++i) {
      for (j = 0; j < n; ++j) {
        ny[i][j] = ksm(a[i] + a[j], 1000000007 - 2);
      }
    }
    for (i = 0; i < (1 << n); ++i) {
      F[i] = 1;
      for (j = 0; j < n; ++j) {
        if ((i >> j) & 1) (F[i] *= a[j]) %= 1000000007, ++sz[i];
      }
    }
    mx = (1 << n) - 1;
    for (i = 1; i < (1 << n); ++i) {
      f[i] = 1;
      x = (i & -i);
      for (j = i & (i - 1); j; j = (j - 1) & i)
        (f[i] -= f[j] * CAL(j, i ^ j)) %= 1000000007;
    }
    dp[0] = 1;
    for (i = 0; i < mx; ++i) {
      f1 = i ^ mx;
      (ans += CAL(f1, i) % 1000000007 * f[f1] % 1000000007 * sz[f1]) %=
          1000000007;
    }
    cout << (ans + 1000000007) % 1000000007;
  }
  