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;
#define MAXNUM 555555
#define rep(i, s, t) for(int i=s;i<t;i++)
set<int> sl,sr;
int num[MAXNUM], pos[MAXNUM], degree[MAXNUM];
vector<int> e[MAXNUM], res,tmp;
int main()
{
    int n, k;
    scanf("%d%d", &n, &k);
    rep(i, 1, n + 1)scanf("%d", &num[i]), pos[num[i]] = i;
    rep(i,1,k + 1)sr.insert(num[i]);
    rep(i,1,n+1)
    {
        if(k==1)break;
        auto iterl=sl.upper_bound(num[i]),iterr=sr.upper_bound(num[i]);
        if(iterl!=sl.end())
            e[pos[*iterl]].push_back(i),degree[i]++;
        if(iterr!=sr.end())
            e[pos[*iterr]].push_back(i),degree[i]++;
        sl.insert(num[i]);
        if(i-k+1>=1)
            sl.erase(num[i-k+1]);
        sr.erase(num[i]);
        if(i+k<=n)
            sr.insert(num[i+k]);
    }
    priority_queue<int> pq;
    rep(i, 1, n + 1)if (!degree[i])pq.push(i);
    rep(i, 1, n + 1)
    {
        int k = pq.top();
        pq.pop();
        tmp.push_back(k);
        for (int kk:e[k])
        {
            degree[kk]--;
            if (!degree[kk])pq.push(kk);
        }
    }
    reverse(tmp.begin(),tmp.end());
    res.resize(n + 1);
    rep(i,0,n)res[tmp[i]]=i+1;
    rep(i,1,res.size())
        printf("%d\n",res[i]);
}