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>
  #pragma GCC optimize("O2,unroll-loops")
  using namespace std;
  const int inf = 1000000010;
  const long long INF = 1000000000000001000LL;
  const int mod = 1000000007;
  const int N = 14;
  int n, m, k, u, v, x, y, t, a, b, ans;
  long long A[N], P[N][N];
  long long tof[N][1 << N];
  long long tof2[1 << 7][1 << N], tof3[1 << 7][1 << N];
  long long dp[1 << N];
  long long powmod(long long a, long long b) {
    long long res = 1;
    for (; b; b >>= 1, a = a * a % mod)
      if (b & 1) res = res * a % mod;
    return res;
  }
  long long Get(int mask1, int mask2) {
    int x = (mask1 & 127), y = ((mask1 - x) >> 7);
    return tof2[x][mask2] * tof3[y][mask2] % mod;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    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++)
        P[i][j] = A[i] * powmod(A[i] + A[j], mod - 2) % mod;
    for (int i = 0; i < n; i++) {
      tof[i][0] = 1;
      for (int j = 1; j < (1 << n); j++) {
        int v = __builtin_ctz(j);
        tof[i][j] = tof[i][j ^ (1 << v)] * P[i][v] % mod;
      }
    }
    for (int i = 0; i < (1 << n); i++) tof2[0][i] = tof3[0][i] = 1;
    for (int i = 1; i < (1 << 7); i++) {
      int v = __builtin_ctz(i);
      for (int mask = 0; mask < (1 << n); mask++) {
        tof2[i][mask] = tof2[i ^ (1 << v)][mask] * tof[v][mask] % mod;
        tof3[i][mask] = tof3[i ^ (1 << v)][mask] * tof[v + 7][mask] % mod;
      }
    }
    dp[0] = 1;
    for (int mask = 1; mask < (1 << n); mask++) {
      dp[mask] = 1;
      for (int sub = (mask - 1) & mask; sub; sub = (sub - 1) & mask)
        dp[mask] = (dp[mask] - dp[sub] * Get(sub, mask ^ sub)) % mod;
      long long ted = __builtin_popcount(mask);
      ans = (ans + dp[mask] * Get(mask, (1 << n) - 1 - mask) % mod * ted) % mod;
    }
    if (ans < 0) ans += mod;
    cout << ans << "\n";
    return 0;
  }
  