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<queue>
#include<cstdio>
#include<cstring>
#define inf 2e8
#define maxn 500010
#define min(x,y) (x)<(y)?(x):(y)
using namespace std;
int to[maxn*2],nxt[maxn*2],mn[maxn*4];
int a[maxn],b[maxn],d[maxn];
int first[maxn];
int n,k,cnt;
priority_queue<int>q;
void add(int x,int y)
{
    to[++cnt]=y;
    nxt[cnt]=first[x];
    first[x]=cnt;
    d[y]++;
}
void change(int x,int l,int r,int p,int v)
{
    if(l==r)
    {
        mn[x]=v;
        return;
    }
    int mid=(l+r)>>1;
    if(p<=mid)
        change(x<<1,l,mid,p,v);
    else
        change(x<<1|1,mid+1,r,p,v);
    mn[x]=min(mn[x<<1],mn[x<<1|1]);
}
int query(int x,int l,int r,int s,int t)
{
    if(s<=l&&r<=t)
        return mn[x];
    int mid=(l+r)>>1,ans=inf;
    if(s<=mid)
        ans=min(ans,query(x<<1,l,mid,s,t));
    if(t>mid)
        ans=min(ans,query(x<<1|1,mid+1,r,s,t));
    return ans;
}
int main()
{
    int x;
    scanf("%d%d",&n,&k);
    memset(mn,15,sizeof(mn));
    for(int i=1;i<=n;i++)
        scanf("%d",&b[i]),a[b[i]]=i;
    for(int i=n;i;i--)
    {
        x=query(1,1,n,a[i]-k+1,a[i]);
        if(x<=n)
            add(a[x],a[i]);
        x=query(1,1,n,a[i],a[i]+k-1);
        if(x<=n)
            add(a[x],a[i]);
        change(1,1,n,a[i],i);
    }
    for(int i=1;i<=n;i++)
        if(!d[i])
            q.push(i);
    for(int i=n;i;i--)
    {
        b[i]=x=q.top();
        q.pop();
        for(int j=first[x];j;j=nxt[j])
            if((--d[to[j]])==0)
                q.push(to[j]);
    }
    for(int i=1;i<=n;i++)
        a[b[i]]=i;
    for(int i=1;i<=n;i++)
        printf("%d\n",a[i]);
    return 0;
}