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>
#define maxn 500010
#define ms(x,y) memset(x,y,sizeof x)
#define inf 0x3f3f3f3f
#define fir first
#define sec second
#define lb(x) (x&(-x))
const int mod=1e9+7;
using namespace std;
typedef long long ll;
set<int>::iterator tmp;
int n,k;
int a[maxn],pos[maxn],deg[maxn];
set<int>front,back;
vector<int>graph[maxn];
int main(){
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)scanf("%d",&a[i]),pos[a[i]]=i;
	for(int i=1;i<k;i++)front.insert(a[i]);
	for(int i=1;i<=n;i++){
		front.erase(a[i-1]);
		if(i+k-1<=n)front.insert(a[i+k-1]);
		back.insert(a[i]);
		if(i-k>=1)back.erase(a[i-k]);
		tmp=front.upper_bound(a[i]);
		if(tmp!=front.end()){
			graph[pos[(*tmp)]].push_back(i);
			deg[i]++;
		}
		tmp=back.upper_bound(a[i]);
		if(tmp!=back.end()){
			graph[pos[(*tmp)]].push_back(i);
			deg[i]++;
		}
	}
	front.clear();
	for(int i=1;i<=n;i++)if(!deg[i])front.insert(-i);
	int top=0;
	while(!front.empty()){
		int u=-(*front.begin());
		a[++top]=u;
		front.erase(front.begin());
		for(int i=0;i<graph[u].size();i++){
			int e=graph[u][i];
			if(--deg[e]==0)front.insert(-e);
		}
	}
	for(int i=1;i<=n;i++)pos[a[i]]=n-i+1;
	for(int i=1;i<=n;i++)printf("%d\n",pos[i]); 
	return 0;
}
