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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <cmath>
#include <map>
using namespace std;
#define MOD 1000000007
#define ADD(X,Y) ((X) = ((X) + (Y)%MOD) % MOD)
typedef long long i64; typedef vector<int> ivec; typedef vector<string> svec;

const int INF = 2100000000;

struct segtree
{
	static const int DEPTH = 19;
	static const int N = 1 << DEPTH;
	int val[2 * N];

	segtree() {
		for (int i = 0; i < 2 * N; ++i) val[i] = INF;
	}
	void update(int p, int v) {
		p += N; val[p] = v; p >>= 1;
		while (p) {
			val[p] = min(val[p * 2], val[p * 2 + 1]);
			p >>= 1;
		}
	}
	int query(int L, int R) {
		L += N; R += N;
		int ret = INF;
		while (L < R) {
			if (L & 1) ret = min(ret, val[L++]);
			if (R & 1) ret = min(ret, val[--R]);
			L >>= 1; R >>= 1;
		}
		return ret;
	}
};

segtree seg;
int N, K, P[505050];
bool visited[505050];
int ans[505050], al;

void decide(int idx)
{
	for (;;) {
		int least = seg.query(0, P[idx]);
		if (least >= idx + K) break;
		decide(least);
	}
	ans[idx] = al++;
	visited[idx] = true;
	seg.update(P[idx], INF);
}

int main()
{
	scanf("%d%d", &N, &K);
	for (int i = 0; i < N; ++i) {
		scanf("%d", P + i); --P[i];
		visited[i] = false;
		seg.update(P[i], i);
	}
	al = 0;
	for (int i = 0; i < N; ++i) if (!visited[i]) decide(i);
	for (int i = 0; i < N; ++i) printf("%d\n", ans[i] + 1);
	return 0;
}
