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 l(x) (x<<1)
#define r(x) (x<<1)|1
#define pb push_back
const int N = 500000 + 10;
using namespace std;

int n,m,e;
int mx[N<<2],t[N],s[N],b[N],bgn[N<<1],to[N<<2],nxt[N<<2];
int v[N],snt;

priority_queue<int, vector<int>, greater<int> > q;
namespace seg
{
  inline int ckmx(int a,int b)
  {
    return a>b?a:b;
  }
  void mt(int x){mx[x]=ckmx(mx[l(x)],mx[r(x)]);}
  void upd(int x,int l,int r,int p,int v)
  {
    if(l==r) {mx[x]=v; return;} 
	int m=(l+r)>>1;
	if(p<=m) upd(l(x),l,m,p,v); else upd(r(x),m+1,r,p,v);
	mt(x);
  }
  int qury(int x,int l,int r,int ql,int qr)
  {
    if(ql<=l&&r<=qr) return mx[x]; 
	else
	{
	  int g=0,m=(l+r)>>1;
	  if(ql<=m) g=ckmx(g,qury(l(x),l,m,ql,qr));
	   if(m<qr) g=ckmx(g,qury(r(x),m+1,r,ql,qr));
	  return g;
	}
  }
}

using namespace seg;

namespace solver
{
  int abs(int x) { return x > 0 ? x : -x;}
  void add(int x,int y)
  {
    to[++e]=y; nxt[e]=bgn[x]; bgn[x]=e;
  }
  void dfs(int x=0)
  {
    int sz=0;
	s[snt++]=x;
    for(int i=bgn[x]; i; i=nxt[i]) q.push(to[i]);
	while(!q.empty())
	{
	  int y=q.top(); q.pop();  dfs(y);
	}
   }
  void solve()
  {
    scanf("%d%d",&n,&m);
    int x(0);
	for(int i=1; i<=n; ++i)
	{
	  scanf("%d",&x);
	  b[i]=x;t[x]=i;
	}	
	for(int i=1; i<=n; ++i)
	{
	  int ps=qury(1,1,n,t[i]+1,t[i]+m-1);
      upd(1,1,n,t[i],i);
	  add(t[ps],t[i]);
	}
	dfs();
	for(int i=1; i<=n; ++i) b[s[i]]=i;
	for(int i=1; i<=n; ++i) printf("%d\n",b[i]);
  }    
}

int main()
{
  solver :: solve();
}