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<vector>
#include<algorithm>
#define eb emplace_back
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef pair<int,int>pi;
typedef pair<ll,ll>pl;
const int inf=1e9+7;
int a[500010],b[500010];
int sm[500010],pm[500010];
int k;
void sort(int s,int e)
{
    if(e-s<2)
        return;
    int m=s+e>>1;
    sort(s,m);
    sort(m,e);
    int p=s,q=m;
    sm[m]=inf;
    pm[m-1]=-inf;
    int i;
    for(i=m;i-->s;)
        sm[i]=min(a[i],sm[i+1]);
    for(i=m;i<e;i++)
        pm[i]=max(a[i],pm[i-1]);
    for(i=s;i<e;i++)
    {
        if(p>=m)
            b[i]=a[q++];
        else if(q>=e)
            b[i]=a[p++];
        else if(sm[p]>=pm[q]+k)
            b[i]=a[q++];
        else
            b[i]=a[p++];
    }
    for(i=s;i<e;i++)
        a[i]=b[i];
    return;
}
int n,i;
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin>>n>>k;
    for(i=0;i<n;i++)
    {
        int s;
        cin>>s;
        a[s-1]=i;
    }
    sort(0,n);
    for(i=0;i<n;i++)
        b[a[i]]=i+1;
    for(i=0;i<n;i++)
        cout<<b[i]<<'\n';
    cout<<endl;
    return 0;
}