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;

template <class I> I chmin(I &x, I y) { return x > y ? x = y : x; }

const int MaxN(500003);
const int INF(0x3f3f3f3f);
int p[MaxN], q[MaxN], d[MaxN];
vector<int> G[MaxN];

#define lc rt << 1
#define rc rt << 1 | 1

int Seg[MaxN << 2];

void Build(int rt, int l, int r)
{
	Seg[rt] = INF;
	if(l == r) return;
	int mid = (l + r) >> 1;
	Build(lc, l, mid);
	Build(rc, mid + 1, r);
}

void Modify(int rt, int l, int r, int pos, int v)
{
	chmin(Seg[rt], v);
	if(l == r) return;
	int mid = (l + r) >> 1;
	if(pos <= mid) Modify(lc, l, mid, pos, v);
	else Modify(rc, mid + 1, r, pos, v);
}

int Query(int rt, int l, int r, int ql, int qr)
{
	if(ql <= l && r <= qr)
		return Seg[rt];
	int mid = (l + r) >> 1;
	int res = INF;
	if(ql <= mid) chmin(res, Query(lc, l, mid, ql, qr));
	if(qr > mid) chmin(res, Query(rc, mid + 1, r, ql, qr));
	return res;
}

#undef lc
#undef rc

int main()
{
	int n, K;
	scanf("%d%d", &n, &K);
	for(int i = 1; i <= n; i++)
	{
		scanf("%d", &p[i]);
		q[p[i]] = i;
	}
	Build(1, 1, n);
	for(int i = n, x; i; i--)
	{
		x = Query(1, 1, n, max(q[i] - K + 1, 1), q[i]);
		if(x != INF) G[q[i]].push_back(q[x]), ++d[q[x]];
		x = Query(1, 1, n, q[i], min(q[i] + K - 1, n));
		if(x != INF) G[q[i]].push_back(q[x]), ++d[q[x]];
		Modify(1, 1, n, q[i], i);
	}
	set<int> Q;
	for(int i = 1; i <= n; i++)
		if(!d[i]) Q.insert(i);
	for(int i = 1; i <= n; i++)
	{
		int u = q[i] = *Q.begin();
		Q.erase(Q.begin());
		for(auto v : G[u])
			if(!--d[v]) Q.insert(v);
	}
	for(int i = 1; i <= n; i++)
		p[q[i]] = i;
	for(int i = 1; i <= n; i++)
		printf("%d\n", p[i]);
	return 0;
}
