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<cstdio>
#include<cstring>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
#define maxn 500005
#define LL long long
using namespace std;
typedef set<int>::iterator siter;
int n,k,p[maxn],ans[maxn],pos[maxn],vis[maxn],pr[maxn],cnt;
set<int> fr,bk;
vector<int> a[maxn];
priority_queue<int> q;
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++) 
	{
		scanf("%d",&p[i]);
		ans[i]=p[i],pos[p[i]]=i;
	}
	for(int i=1;i<=k;i++) bk.insert(p[i]);
	for(int i=1;i<=n;i++)
	{
		siter f=bk.upper_bound(p[i]),g=fr.upper_bound(p[i]);
		if(f!=bk.end())
		{
			a[pos[*f]].push_back(i);
			vis[i]++;
		}
		if(g!=fr.end())
		{
			a[pos[*g]].push_back(i);
			vis[i]++;
		}
		fr.insert(p[i]);
		if(i-k+1>0) fr.erase(p[i-k+1]);
		if(i+k<=n) bk.insert(p[i+k]);
		bk.erase(p[i]);
	}
	for(int i=1;i<=n;i++)
		if(!vis[i]) q.push(i);
	while(!q.empty())
	{
		int x=q.top(); q.pop();
		pr[++cnt]=x;
		for(int i=0,siz=a[x].size();i<siz;i++)
		{
			vis[a[x][i]]--;
			if(!vis[a[x][i]]) q.push(a[x][i]);
		}
	}
	for(int i=1;i<=n;i++) p[pr[i]]=n-i+1;
	for(int i=1;i<=n;i++) printf("%d\n",p[i]);
}