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 <iostream>
#include <set>
using namespace std;
const int N = 510000;
int n, k, ind[N];
int a[N], b[N];
set<int>f, g;
struct Edge {
	int to;
	Edge *next;
}*h[N], e[N<<2];
int _;
inline void add(int u, int v) {
	Edge *tmp = &e[++_];
	tmp -> to = v;
	tmp -> next = h[u];
	h[u] = tmp;
}
inline char nc() {
	static char buf[100000], *p1, *p2;
	return p1==p2&&(p2=(p1=buf)+fread(buf, 1, 100000, stdin)), p1==p2?EOF:*p1++;
}
int read() {
	int x = 0; char ch = nc();
	while(!isdigit(ch)) ch = nc();
	while(isdigit(ch)) x = x*10+ch-48, ch = nc();
	return x;
}
int main() {
	n = read(), k = read();
	for(int i = 1; i <= n; i ++ ) b[a[i]=read()] = i;
	for(int i = 1; i < k; i ++ ) f.insert(a[i]);
	for(int i = 1; i <= n; i ++ ) {
		if(i+k-1 < n) f.insert(a[i+k-1]);
		f.erase(a[i-1]);
		g.insert(a[i]);
		if(i > k) g.erase(a[i-k]);
		set<int>::iterator cur;
		cur = f.upper_bound(a[i]);
		if(cur != f.end())
			add(b[*cur], i), ind[i]++;
		cur = g.upper_bound(a[i]);
		if(cur != g.end()) {
			add(b[*cur], i), ind[i]++;
		}
	}
	f.clear();
	for(int i = 1; i <= n; i ++ ) if(!ind[i]) f.insert(-i);
	int top = 0;
	while(f.size()) {
		int  u = -(*f.begin());
		a[++top] = u;
		f.erase(f.begin());
		for(Edge *i = h[u]; i; i = i->next) {
			if(--ind[i->to] == 0) {
				f.insert(-i->to);
			}
		}
	}
	for(int i = 1; i <= n; i ++ ) {
		b[a[i]] = n-i+1;
	}
	for(int i = 1; i <= n; i ++ ) cout << b[i] << endl;
}
