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;
const int N = 5e5 + 100;
set<int> s;
int h[N], d[N], cnt, n, k, q[N], p[N];
priority_queue<int, vector<int>, greater<int> > r;
struct edge{
	int to, next;
}e[N<<2];

template<class I> inline void gi(I & x) {
	int f = 1;
	char c; x = 0;
	while (!isdigit(c = getchar())) if (c == '-') f = -1;
	while (isdigit(c)) x = x * 10 + (c & 15), c = getchar();
}

void add(int u, int v) {
	e[++cnt] = (edge) {v, h[u]}; h[u] = cnt;
	d[v] ++;
}

int main() {
	gi(n), gi(k);
	for (int i = 1; i <= n; i++) gi(p[i]), q[p[i]] = i;
    set<int>::iterator it;
	for (int i = 1; i <= n; i++) {
		if (i > k) s.erase(p[i - k]);
		it = s.insert(p[i]).first;
		if (++it != s.end()) add(i, q[*it]);
	}
	s.clear();
	for (int i = n; i; i--) {
		if (i + k <= n) s.erase(p[i+k]);
		it = s.insert(p[i]).first;
		if (++it != s.end()) add(i, q[*it]);
	}
	for (int i = 1; i <= n; i++) if (!d[i]) r.push(i);
	for (int i = 1; i <= n; i++) {
		int x = r.top(); r.pop();
		p[x] = i;
		for (int j = h[x]; j; j = e[j].next) {
			int v = e[j].to;
			d[v] --;
			if (!d[v]) r.push(v);
		}
	}
	for (int i = 1; i <= n; i++) printf("%d\n", p[i]);

}


