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>
#include<set>
#include<vector>
#define N 500005
using namespace std;
vector<int> G[N];
set<int> S,T;
set<int> :: iterator it;
int n,k,a[N],b[N],d[N],w[N],m;
int main()
{
	int i,x;
	scanf("%d %d",&n,&k);
	for(i=1;i<=n;i++)
		scanf("%d",&a[i]),b[a[i]]=i;
	for(i=1;i<k;i++) T.insert(a[i]);
	for(i=1;i<=n;i++){
		if(i+k-1<=n) T.insert(a[i+k-1]);
		if(i) T.erase(a[i-1]);
		S.insert(a[i]);
		if(i>k) S.erase(a[i-k]);
		it=S.upper_bound(a[i]);
		if(it!=S.end())
			G[b[*(it)]].push_back(i),d[i]++;
		it=T.upper_bound(a[i]);
		if(it!=T.end())
			G[b[*(it)]].push_back(i),d[i]++;
	  }
	S.clear();
	for(i=1;i<=n;i++)
		if(!d[i]) S.insert(-i);
	while(S.size()){
		x=-(*S.begin()),w[++m]=x;
		S.erase(S.begin());
		for(i=0;i<G[x].size();i++)
			if(!(--d[G[x][i]]))
				S.insert(-G[x][i]);
	  }
	for(i=1;i<=n;i++) b[w[i]]=n-i+1;
	for(i=1;i<=n;i++)
		printf("%d\n",b[i]);
	return 0;
}
