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>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define lb(a) lower_bound(a)
#define is(a) insert(a)
#define ps(a) push(a)
const int N=5e5+10;
const int inf=0x3f3f3f3f;
int in[N],loc[N],p[N],ans[N],tot,n,k;
std::priority_queue <int,std::vector<int>,std::greater<int> > q;
int head[N],to[N<<1],Next[N<<1],cnt;
void add(int u,int v)
{
    to[++cnt]=v,Next[cnt]=head[u],head[u]=cnt;
}
#define ls id<<1
#define rs id<<1|1
int min(int x,int y){return x<y?x:y;}
int max(int x,int y){return x>y?x:y;}
int mi[N<<2];
void change(int id,int l,int r,int p,int d)
{
    if(l==r) {mi[id]=d;return;}
    int mid=l+r>>1;
    if(p<=mid) change(ls,l,mid,p,d);
    else change(rs,mid+1,r,p,d);
    mi[id]=min(mi[ls],mi[rs]);
}
int query(int id,int L,int R,int l,int r)
{
    if(mi[id]==inf) return inf;
    if(l==L&&r==R) return mi[id];
    int Mid=L+R>>1;
    if(r<=Mid) return query(ls,L,Mid,l,r);
    else if(l>Mid) return query(rs,Mid+1,R,l,r);
    else return min(query(ls,L,Mid,l,Mid),query(rs,Mid+1,R,Mid+1,r));
}
int main()
{
    scanf("%d%d",&n,&k);
    rep(i,1,n) scanf("%d",p),p[p[0]]=i;
    memset(mi,0x3f,sizeof(mi));
    dep(i,n,1)
    {
        int loc=query(1,1,n,p[i],min(n,p[i]+k-1));
        if(loc!=inf) add(p[i],p[loc]),++in[p[loc]];
        loc=query(1,1,n,max(1,p[i]-k+1),p[i]);
        if(loc!=inf) add(p[i],p[loc]),++in[p[loc]];
        change(1,1,n,p[i],i);
    }
    rep(i,1,n) if(!in[i]) q.ps(i);
    while(!q.empty())
    {
        int now=q.top();
        ans[now]=++tot;
        q.pop();
        for(int i=head[now];i;i=Next[i])
        {
            int v=to[i];
            --in[v];
            if(!in[v]) q.ps(v);
        }
    }
    rep(i,1,n) printf("%d\n",ans[i]);
    return 0;
}