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 il inline
#define RG register
#define ll long long
#define N (2000010)

using namespace std;

set<int> S;
set<int>::iterator it;
priority_queue<int> Q;

struct edge{ int nt,to; }g[N];

int head[N],p[N],q[N],d[N],k,n,num;

il int gi(){
  RG int x=0,q=1; RG char ch=getchar();
  while ((ch<'0' || ch>'9') && ch!='-') ch=getchar();
  if (ch=='-') q=-1,ch=getchar();
  while (ch>='0' && ch<='9') x=x*10+ch-48,ch=getchar();
  return q*x;
}

il void insert(RG int from,RG int to){
  g[++num]=(edge){head[from],to},head[from]=num,++d[to]; return;
}

int main(){
#ifndef ONLINE_JUDGE
  freopen("permutation.in","r",stdin);
  freopen("permutation.out","w",stdout);
#endif
  n=gi(),k=gi();
  for (RG int i=1;i<=n;++i) q[p[i]=gi()]=i;
  for (RG int i=1;i<=n;++i){
    if (i>k) S.erase(p[i-k]);
    it=S.insert(p[i]).first;
    if (++it!=S.end()) insert(i,q[*it]);
  }
  S.clear();
  for (RG int i=n;i;--i){
    if (i+k<=n) S.erase(p[i+k]);
    it=S.insert(p[i]).first;
    if (++it!=S.end()) insert(i,q[*it]);
  }
  for (RG int i=1;i<=n;++i) if (!d[i]) Q.push(-i);
  for (RG int id=1;id<=n;++id){
    RG int x=-Q.top(); Q.pop(); p[x]=id;
    for (RG int i=head[x];i;i=g[i].nt)
      if (!(--d[g[i].to])) Q.push(-g[i].to);
  }
  for (RG int i=1;i<=n;++i) printf("%d\n",p[i]); return 0;
}
