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 MAXN = 500 * 1000 + 23, INF = 1000 * 1000 * 1000 + 23;










int n, k, p[MAXN], inv[MAXN];

vector<int> kmrg(vector<int> u, vector<int> v) {
	vector<int> m, res;
	reverse(u.begin(), u.end());
	m.push_back(u[0]);
	for (int i = 1; i < (int)u.size(); i++) {
		int tmp = m.back();
		m.push_back(min(u[i], tmp));
	}
	reverse(v.begin(), v.end());
	while (!u.empty() && !v.empty())
		if (v.back() <= m.back() - k) res.push_back(v.back()), v.pop_back();
		else res.push_back(u.back()), m.pop_back(), u.pop_back();
	while (!u.empty()) res.push_back(u.back()), u.pop_back();
	while (!v.empty()) res.push_back(v.back()), v.pop_back();
	return res;
}


vector<int> msrt(int l, int r) {
	if (r - l == 1) return {inv[l]};
	int mid = (l + r) >> 1;
	vector<int> u = msrt(l, mid), v = msrt(mid, r);
	return kmrg(u, v);
}




int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr), cout.tie(nullptr);
	cin >> n >> k;
	for (int i = 0; i < n; i++) {
		cin >> p[i];
		inv[--p[i]] = i;
	}
	vector<int> vec = msrt(0, n);
	for (int i = 0; i < n; i++) p[vec[i]] = i;
	for (int i = 0; i < n; i++) cout << p[i] + 1 << '\n';
	return 0;
}