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<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<vector>
using namespace std;

typedef long long ll;
const int MAXN=5e5+5;

int n,K;
int p[MAXN],q[MAXN],in[MAXN];
set<int> f,b;
set<int>::iterator it;
vector<int> E[MAXN];

int main(){

    scanf("%d%d",&n,&K);
    for(int i=1;i<=n;i++){
        scanf("%d",&p[i]);
        q[p[i]]=i;
    }
    for(int i=1;i<K;i++) f.insert(p[i]);
    for(int i=1;i<=n;i++){
        if(i+K-1<=n) f.insert(p[i+K-1]);
        f.erase(p[i-1]);
        b.insert(p[i]);
        if(i>K) b.erase(p[i-K]);

        it=f.upper_bound(p[i]);
        if(it!=f.end()) E[q[*it]].push_back(i),in[i]++;
        it=b.upper_bound(p[i]);
        if(it!=b.end()) E[q[*it]].push_back(i),in[i]++;
    }
    f.clear();
    for(int i=1;i<=n;i++)
        if(!in[i])
            f.insert(-i);
    int l=0;
    while(f.size()){
        int u=-(*f.begin());
        p[++l]=u;
        f.erase(f.begin());
        for(int i=0;i<(int)E[u].size();i++){
            int v=E[u][i];
            if((--in[v])==0)
                f.insert(-v);
        }
    }
    for(int i=1;i<=n;i++)
        q[p[i]]=n-i+1;
    for(int i=1;i<=n;i++)
        printf("%d\n",q[i]);

    fclose(stdin);
    fclose(stdout);
}
