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 push(x) vec.push_back(x)
const int MN = 501234;
int a[MN], b[MN], c[MN], n, k;

void sort(int l, int r) {
	if (r - l == 1)
		return;
	int mid = (l + r) / 2;
	sort(l, mid);
	sort(mid, r);
	for (int i = mid - 1, mn = MN; i >= l; i--) {
		mn = min(mn, b[i]);
		c[i] = mn;
	}
	int pl = l,	pr = mid, p = l;
	vector <int> vec;
	while (vec.size() != r - l) {
		if (pl == mid)
			push(b[pr++]);
		else if (pr == r)
			push(b[pl++]);
		else if (b[pr] + k <= c[pl])
			push(b[pr++]);
		else
			push(b[pl++]);
	}
	for (int i = l; i < r; i++)
		b[i] = vec[i - l];
}

int main() {
	ios::sync_with_stdio(false);
	cin >> n >> k;
	for (int i = 0; i < n; i++)
		cin >> a[i];
	for (int i = 0; i < n; i++)
		b[a[i] - 1] = i;
	sort(0, n);
	for (int i = 0; i < n; i++)
		a[b[i]] = i + 1;
	for (int i = 0; i < n; i++)
		cout << a[i] << ' ';
}