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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;
int n,k,cnt=-1;
int p[500010],mx[2000010],pos[500010],ans[500010];
vector<int>edge[500010];
void modify(int x,int l,int r,int pos,int v)
{
    if(l==r)
    {
        mx[x]=v;
        return;
    }
    int mid=(l+r)>>1;
    if(pos<=mid)
        modify(x<<1,l,mid,pos,v);
    else modify(x<<1|1,mid+1,r,pos,v);
    mx[x]=max(mx[x<<1],mx[x<<1|1]);
}
int query(int x,int l,int r,int a,int b)
{
    if(a<=l&&r<=b)
        return mx[x];
    int mid=(l+r)>>1,res=0;
    if(a<=mid)
        res=max(res,query(x<<1,l,mid,a,b));
    if(b>mid)
        res=max(res,query(x<<1|1,mid+1,r,a,b));
    return res;
}
void dfs(int x)
{
    ans[++cnt]=x;
    sort(edge[x].begin(),edge[x].end());
    for(int i=0; i<(int)edge[x].size(); i++)
        dfs(edge[x][i]);
}
inline int read()
{
    int x=0;
    char ch=getchar();
    while(ch<'0'||ch>'9')
        ch=getchar();
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x;
}
int main()
{
    n=read(),k=read();
    for(int i=1; i<=n; i++)
    {
        p[i]=read();
        pos[p[i]]=i;
    }
    pos[n+1]=n+1;
    for(int i=1; i<=n; i++)
    {
        int id=query(1,1,n,pos[i]+1,pos[i]+k-1);
        if(!id)
            id=n+1;
        edge[pos[id]].push_back(pos[i]);
        modify(1,1,n,pos[i],i);
    }
    dfs(n+1);
    for(int i=1; i<=n; i++)
        p[ans[i]]=i;
    for(int i=1; i<=n; i++)
        printf("%d\n",p[i]);
    return 0;
}
