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 N = 500005;
int n, m, a[N];
int seg[N * 4];
void update(int k, int x, int i = 1, int l = 1, int r = n) {
	if (l == r) {
		seg[i] = x;
		return;
	}
	int mid = (l + r) >> 1;
	if (k <= mid)
		update(k, x, i << 1, l, mid);
	else
		update(k, x, i << 1 | 1, mid + 1, r);
	seg[i] = max(seg[i << 1], seg[i << 1 | 1]);
}
int query(int lf, int rg, int i = 1, int l = 1, int r = n) {
	if (lf > rg) return 0;
	if (lf <= l && r <= rg)
		return seg[i];
	int res = 0;
	int mid = (l + r) >> 1;
	if (lf <= mid)
		res = max(res, query(lf, rg, i << 1, l, mid));
	if (rg > mid)
		res = max(res, query(lf, rg, i << 1 | 1, mid + 1, r));
	return res;
}
int ind[N], lst[N], ans[N];
vector<int> G[N];
priority_queue<int> que;
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		int x;
		scanf("%d", &x);
		a[x] = i;
	}
	for (int i = 1; i <= n; i++) {
		int j;
		j = query(max(1, a[i] - m + 1), a[i] - 1);
		if (j) G[a[i]].push_back(a[j]), ind[a[j]]++;
		j = query(a[i] + 1, min(n, a[i] + m - 1));
		if (j) G[a[i]].push_back(a[j]), ind[a[j]]++;
		update(a[i], i);
	}
	for (int u = 1; u <= n; u++)
		if (!ind[u])
			que.push(u);
	int cnt = 0;
	while (!que.empty()) {
		int u = que.top();
		lst[++cnt] = u;
		que.pop();
		for (int i = 0; i < (int)G[u].size(); i++) {
			int v = G[u][i];
			ind[v]--;
			if (!ind[v])
				que.push(v);
		}
	}
	reverse(lst + 1, lst + 1 + n);
	for (int i = 1; i <= n; i++)
		ans[lst[i]] = i;
	for (int i = 1; i <= n; i++)
		printf("%d\n", ans[i]);
	return 0;
}