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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<queue>
#include<set>
#define MN 500000
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int n,K,a[MN+5],b[MN+5],head[MN+5],cnt=0,in[MN+5],c[MN+5],d[MN+5];
struct edge{int to,next;}e[MN*4+5];
inline void ins(int f,int t)
{
    if(f<t) swap(f,t);
    f=b[f];t=b[t];++in[t];
    e[++cnt]=(edge){t,head[f]};head[f]=cnt;
}
priority_queue<int> q;set<int> st;
int main()
{
    n=read();K=read();
    for(int i=1;i<=n;++i) b[a[i]=read()]=i;
    for(int i=1;i<n+K;++i)
    {
      //  cout<<i<<"begin\n";
        set<int>::iterator it,it2;
        if(i<=n)
        {
        //    cout<<"Insert"<<a[i]<<endl;
            st.insert(a[i]);
            set<int>::iterator it=st.find(a[i]),it2;
            if(it!=st.begin()) ins(a[i],*(--(it2=it)));
            if((++(it2=it))!=st.end()) ins(a[i],*it2);
        }
        if(i>=K)
        {
          //  cout<<"erase"<<a[i-K+1]<<endl;
            it=st.find(a[i-K+1]);
            if(it!=st.begin()) ins(a[i-K+1],*(--(it2=it)));
            if((++(it2=it))!=st.end()) ins(a[i-K+1],*it2);
            st.erase(it);
        }
    }
    for(int i=1;i<=n;++i) if(!in[i]) q.push(i);
    for(int i=1;i<=n;++i)
    {
        int x=q.top();c[n+1-i]=x;q.pop();
        for(int i=head[x];i;i=e[i].next)
            if(!--in[e[i].to]) q.push(e[i].to);
    }
    for(int i=1;i<=n;++i) d[c[i]]=i;
    for(int i=1;i<=n;++i) printf("%d\n",d[i]);
    return 0;
}
