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 <iostream>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;

int N, K;
vector<int> P, Q;

void mergesort(vector<int> &a, int left, int right) {
	if (right - left <= 1) return;
	int mid = (left + right) / 2;
	mergesort(a, left, mid);
	mergesort(a, mid, right);

	vector<pair<int, int> > rs;
	for (int i = mid; i < right; ++i) rs.push_back({ a[i], i - mid });
	sort(rs.begin(), rs.end(), greater<pair<int,int> >());

	vector<int> pl(right - mid);
	int pointer = mid;
	for (auto pa : rs) {
		int val = pa.first;
		int id = pa.second;
		while (pointer > left && a[pointer - 1] >= val + K) --pointer;
		pl[id] = pointer;
	}
	vector<int> res;
	int last = left;
	for (int i = 0; i < pl.size(); ++i) {
		for (int j = last; j < pl[i]; ++j) {
			res.push_back(a[j]);
		}
		last = max(last, pl[i]);
		res.push_back(a[i + mid]);
	}
	for (int i = last; i < mid; ++i) res.push_back(a[i]);
	for (int i = 0; i < res.size(); ++i) a[i + left] = res[i];
}

int main() {
	cin >> N >> K;
	P.resize(N); Q.resize(N);
	for (int i = 0; i < N; ++i) cin >> P[i], --P[i];
	for (int i = 0; i < N; ++i) Q[P[i]] = i;
	mergesort(Q, 0, N);
	vector<int> res(N);
	for (int i = 0; i < N; ++i) res[Q[i]] = i;
	for (int i = 0; i < N; ++i) cout << res[i]+1 << endl;
}
