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
//Be Name Khoda, Besmellah Rahmane Rahim, In The Name Of God;
#include<bits/stdc++.h>
#include<iostream>
#include<vector>
#include<algorithm>
#include<cmath>
#include<set>
#include<queue>
#include<deque>
#include<map>
#include<stack>
#include<bitset>
#include<list>
#include<cassert>
#include<numeric>
#include<stdio.h>
#include<string.h>
#include<iomanip>
#include<unordered_map>
#include<unordered_set>
#include<iostream>
#include<fstream>

using namespace std;

const int N = 5e5 + 5;
const int inf = N;

int n, k, cnt;
int a[N];
int ans[N];
int ind[N];
int fen[N * 4];
bool mark[N];
priority_queue<int> s;

void add(int ind, int val) 
{
	for (fen[ind += n] = val; ind >>= 1;)
	{
		fen[ind] = max(fen[ind << 1], fen[ind << 1 | 1]);
	}
}

int get(int l, int r) 
{
	l = max(l, 0);
	r = min(r, n);
	int ans = 0;
	for (l += n, r += n; l < r; l >>= 1, r >>= 1) 
	{
		if (l & 1)
		{
			ans = max(ans, fen[l++]);
		}
		if (r & 1)
		{
			ans = max(ans, fen[--r]);
		}
	}
	return ans;
}

void isval(int x) 
{
	int mx = get(x * k, x * k + k);
	if (mx == -inf)
	{
		return;
	}
	int me = ind[mx];
	if (a[me] >= get(me - k + 1, me + k) && mark[me] == false) 
	{
		mark[me] = true;
		s.push(me);
	}
}

int main() 
{
	cin >> n >> k;
	for (int i = 0; i < n; i++) 
	{
		cin >> a[i];
		ind[a[i]] = i;
		add(i, a[i]);
	}
	cnt = (n + k - 1) / k;
	for (int i = 0; i < cnt; i++)
	{
		isval(i);
	}
	for (int i = n; i; i--) 
	{
		int x = s.top();
		s.pop();
		ans[x] = i;
		add(x, -inf);
		x /= k;
		if (x)
		{
			isval(x - 1);
		}
		isval(x);
		if (x + 1 < cnt)
		{
			isval(x + 1);
		}
	}
	for (int i = 0; i < n; i++)
	{
		cout << ans[i] << ' ';
	}
}

