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=500010;
int n,k,a[N],p[N];
int T[N<<1];
int idx(int l,int r){return l+r|l!=r;}
void update(int l,int r,int p,int num){
    int x=idx(l,r),mid=(l+r)>>1;
    if(l==r){T[x]=num;return;}
    if(p<=mid)update(l,mid,p,num);
    else update(mid+1,r,p,num);
    T[x]=max(T[idx(l,mid)],T[idx(mid+1,r)]);
}
int query(int l,int r,int L,int R){
    int x=idx(l,r),mid=(l+r)>>1,res=0;
    if(L<=l&&r<=R)return T[x];
    if(L<=mid)res=max(res,query(l,mid,L,R));
    if(mid<R)res=max(res,query(mid+1,r,L,R));
    return res;
}
int d[N];
vector<int> v[N];
void topsort(){
    set<int> s;
    for(int i=1;i<=n;i++)if(!d[i])s.insert(i);
    for(int i=1;i<=n;i++){
        int x=p[i]=*s.begin(); s.erase(s.begin());
        for(auto y:v[x])if(!(--d[y]))s.insert(y);
    }
}
int main(){
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        p[a[i]]=i;
    }
    for(int i=1;i<=n;i++){
        int t=query(1,n,max(1,p[i]-k+1),p[i]);
        if(t)v[p[t]].push_back(p[i]),d[p[i]]++;
        t=query(1,n,p[i],min(n,p[i]+k-1));
        if(t)v[p[t]].push_back(p[i]),d[p[i]]++;
        update(1,n,p[i],i);
    }
    topsort();
    for(int i=1;i<=n;i++)a[p[i]]=i;
    for(int i=1;i<=n;i++)printf("%d\n",a[i]);
    return 0;
}