#include <bits/stdc++.h>
using namespace std;
long long count(vector<int> c, int L){
  int N = c.size();
  int F = 1;
  for (int i = 0; i < N - 1; i++){
    F *= i + 1;
  }
  long long ans = 0;
  for (int i = 0; i < (1 << N); i++){
    int L2 = L;
    for (int j = 0; j < N; j++){
      if ((i >> j & 1) == 1){
        L2 -= c[j];
      }
    }
    if (L2 >= 0){
      long long p = 1;
      for (int j = 0; j < N - 1; j++){
        p *= L2 + j + 1;
      }
      p /= F;
      if (__builtin_parity(i) == 0){
        ans += p;
      } else {
        ans -= p;
      }
    }
  }
  return ans;
}
int main(){
  int N, L;
  cin >> N >> L;
  vector<int> c(4, 1);
  for (int i = 4 - N; i < 4; i++){
    cin >> c[i];
    c[i]++;
  }
  long long sum = count(c, L);
  int Q;
  cin >> Q;
  for (int i = 0; i < Q; i++){
    long long k;
    cin >> k;
    k--;
    if (k >= sum){
      cout << -1 << endl;
    } else {
      int L2 = L;
      vector<int> ans(4, 0);
      for (int j = 0; j < 4; j++){
        int p = 0;
        if (L2 < c[j]){
          if (k == 0){
            ans[j] = L2;
            break;
          }
          k--;
          p = 1;
        }
        int tv = min(L2, c[j]);
        int fv = 0;
        while (tv - fv > 1){
          int mid = (tv + fv) / 2;
          long long cnt;
          if (j == 0){
            cnt = sum - count(vector<int>{mid, c[1], c[2], c[3]}, L2) - p;
          }
          if (j == 1){
            cnt = count(vector<int>{c[1], c[2], c[3]}, L2) - count(vector<int>{mid, c[2], c[3]}, L2) - p;
          }
          if (j == 2){
            cnt = count(vector<int>{c[2], c[3]}, L2) - count(vector<int>{mid, c[3]}, L2) - p;
          }
          if (cnt <= k){
            tv = mid;
          } else {
            fv = mid;
          }
        }
        ans[j] = fv;
        if (j == 0){
          k -= sum - count(vector<int>{tv, c[1], c[2], c[3]}, L2) - p;
        }
        if (j == 1){
          k -= count(vector<int>{c[1], c[2], c[3]}, L2) - count(vector<int>{tv, c[2], c[3]}, L2)- p;
        }
        if (j == 2){
          k -= count(vector<int>{c[2], c[3]}, L2) - count(vector<int>{tv, c[3]}, L2) - p;
        }
        L2 -= fv;
      }
      for (int j = 4 - N; j < 4; j++){
        cout << ans[j];
        if (j < 3){
          cout << ' ';
        }
      }
      cout << endl;
    }
  }
}