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>
#pragma GCC coptimize("Ofast")
using namespace std;
int n, k;
int site[500007], a[500007];
std::queue<int> q;
bool inq[500007];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    site[a[i]] = i;
    q.push(a[i]);
    inq[a[i]] = 1;
  }
  while (!q.empty()) {
    int i = q.front();
    q.pop();
    inq[i] = 0;
    if (i != n && site[i] - site[i + 1] >= k) {
      swap(a[site[i]], a[site[i + 1]]);
      swap(site[i], site[i + 1]);
      if (!inq[i + 1]) {
        q.push(i + 1);
      }
    }
    if (i != 1 && site[i - 1] - site[i] >= k) {
      swap(a[site[i - 1]], a[site[i]]);
      swap(site[i - 1], site[i]);
      q.push(i);
    }
  }
  for (int i = 1; i <= n; i++) printf("%d\n", a[i]);
}
