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 mid ((s + e) / 2)
#define makefast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define comp(x) lower_bound(vc.begin(), vc.end(), x) - vc.begin()

const int N = 5e5 + 7;

int n, k;
int p[N], q[N];
int seg[N * 50];
bool mrk[N];
int cpk;
set <int> s;
void upd(int p, int val) {
	p += (1 << 20);
	seg[p] = val;
	p /= 2;
	while (p) {
		seg[p] = max(seg[p * 2], seg[p * 2 + 1]);
		p /= 2;
	}
}
int get(int l, int r) {
	l += (1 << 20);
	r += (1 << 20);
	int res = max(seg[l], seg[r]);
	while (l / 2 != r / 2) {
		if (l % 2 == 0)
			res = max(res, seg[l + 1]);
		if (r % 2 == 1)
			res = max(res, seg[r - 1]);
		l /= 2;
		r /= 2;
	}
	return res;
}

void get(int pk) {
	int l = pk * k;
	int r = min(n, pk * k + k - 1);
	int mx = get(l, r);
	if (mx < 0 || mrk[q[mx]])
		return;
	if (get(q[mx] - k + 1, q[mx] + k - 1) == mx) {
		s.insert(q[mx]);
		mrk[q[mx]] = 1;	
	}

}
int res[N];
int main() {
	makefast;
	cin >> n >> k;
	cpk = (n + k - 1) / k;
	for (int i = 0; i < n; i++) {
		cin >> p[i];
		p[i]--;
		q[p[i]] = i;
		upd(i, p[i]);
	}
	for (int i = 0; i < cpk; i++)
		get(i);
	for (int i = 1; i <= n; i++) {
		int p = *s.rbegin();
		s.erase(p);
		res[p + 1] = n - i + 1;
		upd(p, -1);
		for (int i = (p / k) - 1; i <= (p / k) + 1; i++)
			if (0 <= i && i < cpk)
				get(i);
	}
	for (int i = 1; i <= n; i++)
		cout << res[i] << '\n';


	return 0;
}
