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>
using namespace std;
const int N=2e6+5;
int p[N],a[N],b[N];
struct node{int l,r,mid,mn;}s[N<<2];
void build(int di,int l,int r)
{
    s[di].l=l,s[di].r=r,s[di].mid=l+r>>1,s[di].mn=1e9;
    if(l==r)return;
    build(di<<1,l,s[di].mid),build(di<<1|1,s[di].mid+1,r);
}
void ins(int di,int x,int v)
{
    if(s[di].l==s[di].r){s[di].mn=v;return;}
    if(x<=s[di].mid)ins(di<<1,x,v);
    else ins(di<<1|1,x,v);
    s[di].mn=min(s[di<<1].mn,s[di<<1|1].mn);
}
int qry(int di,int l,int r)
{
    if(l<=s[di].l&&s[di].r<=r)return s[di].mn;
    else if(r<=s[di].mid)return qry(di<<1,l,r);
    else if(l>s[di].mid)return qry(di<<1|1,l,r);
    return min(qry(di<<1,l,r),qry(di<<1|1,l,r));
}
struct edge{int to,nxt;}e[N<<1];
int head[N],tot=1;
void adde(int x,int y){e[tot]=(edge){y,head[x]};head[x]=tot++;}
priority_queue<int>q;
int n,rd[N];
void topo()
{
    for(int i=1;i<=n;i++)
        for(int j=head[i];j;j=e[j].nxt)rd[e[j].to]++;
    for(int i=1;i<=n;i++)if(!rd[i])q.push(i);
    int tt=0;
    while(!q.empty())
    {
        int u=q.top();q.pop(),b[++tt]=u;
        for(int i=head[u];i;i=e[i].nxt)
            if(!(--rd[e[i].to]))q.push(e[i].to);
    }
}
int main()
{
    int k;scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){int x;scanf("%d",&x),a[x]=i;}
    build(1,1,n);
    for(int i=n;i>=1;i--)
    {
        int x=qry(1,max(1,a[i]-k+1),a[i]);
        if(x>=1&&x<=n)adde(a[x],a[i]);
        x=qry(1,a[i],min(n,a[i]+k-1));
        if(x>=1&&x<=n)adde(a[x],a[i]);
        ins(1,a[i],i);
    }
    topo();
    for(int i=1;i<=n;i++)p[b[i]]=i;
    for(int i=1;i<=n;i++)printf("%d\n",p[i]);
}