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, inF = N;

int n, k, cnt, a[N], b[N], dex[N];
int seg[N << 1];
bool mark[N];
priority_queue<int> pq;

void change(int dex, int val) {
	for (seg[dex += n] = val; dex >>= 1;)
		seg[dex] = max(seg[dex << 1], seg[dex << 1 | 1]);
}
int get(int l, int r) {
	l = max(l, 0);
	r = min(r, n);

	int res = 0;
	for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
		if (l & 1)
			res = max(res, seg[l++]);
		if (r & 1)
			res = max(res, seg[--r]);
	}
	return res;
}

void check(int x) {
	int mx = get(x * k, x * k + k);
	if (mx == -inF)
		return;
	int me = dex[mx];
	if (a[me] >= get(me - k + 1, me + k) && mark[me] == false) {
		mark[me] = true;
		pq.push(me);
	}
}

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	cin >> n >> k;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		dex[a[i]] = i;
		change(i, a[i]);
	}
	cnt = (n + k - 1) / k;
	for (int i = 0; i < cnt; i++)
		check(i);
	for (int i = n; i; i--) {
		int x = pq.top();
		pq.pop();
		b[x] = i;
		change(x, -inF);
		x /= k;
		if (x)
			check(x - 1);
		check(x);
		if (x + 1 < cnt)
			check(x + 1);
	}
	for (int i = 0; i < n; i++)
		cout << b[i] << ' ';
}
