Problem:
Score : 2000 points

Problem Statement

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.

Sample Input 1

4 2
4 2 3 1

Sample Output 1

2
1
4
3

One possible way to obtain the lexicographically smallest permutation is shown below:

4 2 3 1

4 1 3 2

3 1 4 2

2 1 4 3

Sample Input 2

5 1
5 4 3 2 1

Sample Output 2

1
2
3
4
5

Sample Input 3

8 3
4 5 7 8 3 1 2 6

Sample Output 3

1
2
6
7
5
3
4
8

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 10, inf = 1e8;
int n, nn, k, a[N], b[N], cnt, ans[N];
bool mark[N];
int seg[N << 2];
priority_queue<int> s;
void upd(int p, int val) {
	p += nn, seg[p] = val, p /= 2;
	for (; p > 0; p /= 2)
		seg[p] = max(seg[p * 2], seg[p * 2 + 1]);
	return;
}
int get(int l, int r) {
	int rtr = -inf;
	for (l += nn, r += nn; l < r; l /= 2, r /= 2) {
		if (l % 2) rtr = max(rtr, seg[l++]);
		if (r % 2) rtr = max(rtr, seg[--r]);
	}
	return rtr;
}
void add(int plc) {
	if (plc >= cnt || plc < 0)
		return;
	int mx = get(k * plc, k * (plc + 1));
//	cout << "plc = " << plc << " : " << mx << endl;
	if (mx < 0)
		return;
	if (!mark[b[mx]] && get(max(0, b[mx] - k + 1), min(n, b[mx] + k)) == mx)
		s.push(b[mx]), mark[b[mx]] = true;
//	cout << "done" << endl;
	return;
}
void makeFind(int plc) {
	add(plc - 1);
	add(plc);
	add(plc + 1);
}
int main() {
	ios_base::sync_with_stdio (false), cin.tie(0), cout.tie(0);
	cin >> n >> k, nn = n, cnt = (n + k - 1) / k;
	while (__builtin_popcount(nn) != 1) nn++;
	for (int i = 0; i < n; i++) {
		cin >> a[i], a[i]--, b[a[i]] = i;
		upd(i, a[i]);
	}
	for (int i = 0; i < cnt; i++)
		add(i);
//	cout << "GO NEXT" << endl;
	for (int i = 0; i < n; i++) {
		int ls = s.top();
		s.pop();
		ans[ls] = n - i - 1;
		upd(ls, -inf);
		int plc = ls / k;
		makeFind(plc);
	}
	for (int i = 0; i < n; i++)
		cout << ans[i] + 1 << "\n";
	return 0;
}