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], fen[MAXN], inv[MAXN], od[MAXN], seg[MAXN << 2], ml[MAXN << 2];

inline void upd(int p) { for (p++; p <= MAXN; p += (p & -p)) fen[p]++; }
inline int pget(int p) {
	int ret = 0;
	for (; p; p -= (p & -p)) ret += fen[p];
	return ret;
}
inline int fget(int l, int r) { return pget(min(r, n)) - pget(max(l, 0)); }

void add(int l, int r, int val, int v = 1, int st = 0, int ed = n) {
	if (l >= ed || st >= r) return;
	if (l <= st && ed <= r) {
		seg[v] += val;
		ml[v] += val;
		return;
	}
	int mid = ((st + ed) >> 1);
	add(l, r, val, v << 1, st, mid), add(l, r, val, (v << 1) | 1, mid, ed);
	seg[v] = min(seg[v << 1], seg[(v << 1) | 1]) + ml[v];
}
int get(int v = 1, int st = 0, int ed = n) {
	if (ed - st == 1) return st;
	int mid = ((st + ed) >> 1);
	if (seg[(v << 1) | 1] <= seg[v << 1]) return get((v << 1) | 1, mid, ed);
	else return get(v << 1, st, mid);
}




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;
	}
	for (int i = n - 1; ~i; i--) {
		od[inv[i]] = fget(inv[i] - k + 1, inv[i] + k);
		upd(inv[i]);
	}
	for (int i = 0; i < n; i++) add(i, i + 1, od[i]);
	for (int i = n - 1; ~i; i--) {
		int x = get();
		p[x] = i;
		add(x, x + 1, INF);
		add(max(x - k + 1, 0), min(n, x + k), -1);
	}
	for (int i = 0; i < n; i++) cout << p[i] + 1 << '\n';
	return 0;
}