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: Accepted)

#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 10, inf = 1e8;
int b[N], a[N], n, k, fen[N], val, seg[N << 2], lazy[N << 2], ans[N];
// fen
void add(int p) {
	for (p++; p <= n; p += p & -p)
		fen[p]++;
}
int getPref(int p) {
	int rtr = 0;
	for (; p > 0; p -= p & -p)
		rtr += fen[p];
	return rtr;
}
int get(int l, int r) {
	return getPref(r) - getPref(l);
}
// fen

// seg
void relax(int id) {
	int lc = id * 2, rc = lc + 1;
	seg[lc] += lazy[id], seg[rc] += lazy[id];
	lazy[lc] += lazy[id], lazy[rc] += lazy[id];
	lazy[id] = 0;
}
void addseg(int l, int r, int id = 1, int s = 0, int e = n) {
	if (e <= l || r <= s)
		return;
	if (l <= s && e <= r) {
		seg[id] += val;
		lazy[id] += val;
		return;
	}
	int mid = (s + e) / 2, lc = id * 2, rc = lc + 1;
	relax(id);
	addseg(l, r, lc, s, mid);
	addseg(l, r, rc, mid, e);
	seg[id] = min(seg[lc], seg[rc]);
}
int getseg(int id = 1, int s = 0, int e = n) {
	if (e - s == 1) 
		return s;
	relax(id);
	int mid = (s + e) / 2, lc = id * 2, rc = lc + 1;
	if (seg[rc] == 0)
		return getseg(rc, mid, e);
	return getseg(lc, s, mid);
}
// seg

// main
int main() {
	// input
	ios_base::sync_with_stdio (false), cin.tie(0), cout.tie(0);
	cin >> n >> k;
	for (int i = 0; i < n; i++) 
		cin >> b[i], b[i]--;
	for (int i = 0; i < n; i++)
		a[b[i]] = i;
	// input

	// getDegree
	for (int i = n - 1; i >= 0; i--) {
		val = get(max(0, a[i] - k + 1), min(n, a[i] + k));
		addseg(a[i], a[i] + 1);
		add(a[i]);
	}
	// getDegree

	// get most right ziro
	for (int i = 0; i < n; i++) {
		int plc = getseg();
		ans[plc] = n - i - 1;
		val = -1;
		addseg(max(0, plc - k + 1), min(n, plc + k));
		val = inf;
		addseg(plc, plc + 1);
	}
	// getmost right ziro

	//outout
	for (int i = 0; i < n; i++) 
		cout << ans[i] + 1 << "\n";
	//output

	return 0;
}
// main