You are given a permutation P_1 ... P_N of the set {1, 2, ..., N}.

You can apply the following operation to this permutation, any number of times (possibly zero):

* Choose two indices i,j (1 ≦ i < j ≦ N), such that j - i ≧ K and |P_i - P_j| = 1. Then, swap the values of P_i and P_j.



Among all permutations that can be obtained by applying this operation to the given permutation, find the lexicographically smallest one.

Constraints

* 2≦N≦500,000
* 1≦K≦N-1
* P is a permutation of the set {1, 2, ..., N}.

Input

The input is given from Standard Input in the following format:


N K
P_1 P_2 ... P_N


Output

Print the lexicographically smallest permutation that can be obtained.

Examples

Input

4 2
4 2 3 1


Output

2
1
4
3


Input

5 1
5 4 3 2 1


Output

1
2
3
4
5


Input

8 3
4 5 7 8 3 1 2 6


Output

1
2
6
7
5
3
4
8
#include <bits/stdc++.h>
using namespace std;
const int SIZE = 1e5 + 1;
vector<int> num_arr(SIZE), idx_arr(SIZE);
stack<int> s, t;
int memo[SIZE];
int n, k;
void NIT() {
  for (int i = 1; i <= n; ++i) idx_arr[num_arr[i]] = i;
}
void INT() {
  for (int i = 1; i <= n; ++i) {
    num_arr[idx_arr[i]] = i;
  }
}
void divide_and_conquer_method(int left, int right) {
  if (left >= right) return;
  int median = (left + right) / 2;
  divide_and_conquer_method(left, median);
  divide_and_conquer_method(median + 1, right);
  int Min = idx_arr[median];
  s.push(Min);
  t.push(idx_arr[median]);
  for (int i = median - 1; i >= left; --i) {
    Min = min(Min, idx_arr[i]);
    s.push(Min);
    t.push(idx_arr[i]);
  }
  int j = left;
  for (int i = median + 1; i <= right; i++) {
    while (!(s.empty()) && s.top() - idx_arr[i] < k) {
      memo[j] = t.top();
      ++j;
      s.pop();
      t.pop();
    }
    memo[j] = idx_arr[i];
    j++;
  }
  while (!s.empty()) {
    memo[j] = s.top();
    ++j;
    s.pop();
  }
  for (int i = left; i <= right; i++) {
    idx_arr[i] = memo[i];
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> num_arr[i];
  NIT();
  divide_and_conquer_method(1, n);
  INT();
  for (int i = 1; i <= n; ++i) {
    cout << num_arr[i] << endl;
  }
}
