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 pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define pf push_front
#define pb2 pop_back
#define pf2 pop_front
#define line printf("\n")
#define pq priority_queue
#define rep(k,i,j) for(int k = (int)i;k<(int)j;k++)
#define repd(k,i,j) for(int k = (int)i;k>=(int)j;k--)
#define ll long long
#define ALL(a) a.begin(),a.end()

using namespace std;

double EPS = 1e-9;
int INF = 1e9+7;;
long long INFLL = 1e17;
double pi = acos(-1);
int dirx[8] = {-1,0,0,1,-1,-1,1,1};
int diry[8] = {0,1,-1,0,-1,1,-1,1};

clock_t first_attempt = clock();
inline void cek_time(){
	clock_t cur = clock()- first_attempt;
	cerr<<"TIME : "<<(double) cur/CLOCKS_PER_SEC<<endl;
}
inline void OPEN (string s) {
  freopen ((s + ".in").c_str (), "r", stdin);
  freopen ((s + ".out").c_str (), "w", stdout);
}

//end of template

const int maxn = 5e5+5;
int n,K;

int pos[maxn],ans[maxn],ret[maxn];

int main(){
	scanf("%d%d",&n,&K);
	
	assert(n<=8000);
	rep(k,1,n+1){
		int a;
		scanf("%d",&a);
		pos[a] = k;
	}
	
	repd(k,n-1,1){
		rep(i,k,n){
			if(pos[i]>pos[i+1]-k)break;
			swap(pos[i],pos[i+1]);
		}
	}
	
	rep(k,1,N+1)ans[pos[k]] = k;
	
	rep(k,1,N+1)printf("%d\n",ans[k]);
	return 0;
}