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;

#define ge getchar()
#define Re read()

inline int read() {
	int x = 0, ch;
	while(!isdigit(ch = ge)) ;
	while(isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = ge;
	return x;
}
template<typename T>
inline void chkmin(T&x, T y) { x = x < y ? x : y; }

const int MAXN = 500000;

int a[MAXN + 1];
int d[MAXN + 1];
int to[MAXN + 1][2];
int T[MAXN << 2];
int id[MAXN + 1];

priority_queue<int, vector<int>, greater<int> >q;

#define ls (nd << 1)
#define rs (nd << 1 | 1)

inline void Modify(int nd, int l, int r, int p, int v) {
	T[nd] = v; if(l == r) return ; int mid = (l + r) >> 1;
	if(p <= mid) Modify(ls, l, mid, p, v);
	else Modify(rs, mid + 1, r, p, v);
}

inline int Query(int nd, int l, int r, int L, int R) {
	if(L <= l && r <= R) return T[nd];
	int mid = (l + r) >> 1, res = 2147483647;
	if(L <= mid) chkmin(res, Query(ls, l, mid, L, R));
	if(mid < R) chkmin(res, Query(rs, mid + 1, r, L, R));
	return res;
}

#undef ls
#undef rs

int main() {
	int n = Re, k = Re;
	for(int i = 1; i <= n; i++) a[Re] = i;
	memset(T, 63, sizeof(T));
	for(int i = n; i >= 1; i--) {
		int L = Query(1, 1, n, max(1, a[i] - k + 1), a[i]);
		int R = Query(1, 1, n, a[i], min(n, a[i] + k - 1));
		if(L <= n && a[i] - a[L] < k) d[a[L]]++, to[a[i]][0] = a[L];
		if(R <= n && a[R] - a[i] < k) d[a[R]]++, to[a[i]][1] = a[R];
		Modify(1, 1, n, a[i], i);
	} int x, u, ct = 0;
	for(int i = 1; i <= n; i++) if(!d[i]) q.push(i);
	while(!q.empty()) {
		id[x = q.top()] = ++ct; q.pop();
		u = to[x][0]; if(u) { d[u]--; if(!d[u]) q.push(u); }
		u = to[x][1]; if(u) { d[u]--; if(!d[u]) q.push(u); }
	} for(int i = 1; i <= n; i++) printf("%d\n", id[i]);
	return 0;
}