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 = 5e5 + 5, inf = 0x3f3f3f3f;
int n, k;
int mn[maxn * 4], p[maxn], in[maxn], ans[maxn];
vector<int> G[maxn];
int read() {
	int x = 0, f = 1;
	char c = getchar();
	while(!isdigit(c)) {
		if(c == '-') {
			f = -1;
		}
		c = getchar();
	}
	while(isdigit(c)) {
		x = x * 10 + c - '0';
		c = getchar();
	}
	return x * f;
}
void update(int l, int r, int x, int p, int d) {
	if(l == r) {
		mn[x] = d;
		return;
	}
	int mid = (l + r) >> 1;
	if(p <= mid) {
		update(l, mid, x << 1, p, d);
	} else {
		update(mid + 1, r, x << 1 | 1, p, d);
	}
	mn[x] = min(mn[x << 1], mn[x << 1 | 1]);
}
int query(int l, int r, int x, int a, int b) {
	if(l > b || r < a) {
		return inf;
	}
	if(l >= a && r <= b) {
		return mn[x];
	}
	int mid = (l + r) >> 1;
	return min(query(l, mid, x << 1, a, b), query(mid + 1, r, x << 1 | 1, a, b));
}
int main() {
	n = read();
	k = read();
	memset(mn, inf, sizeof(mn));
	for(int i = 1; i <= n; ++i) {
		p[read()] = i;
	}
	for(int i = n; i; --i) {
		int t = query(1, n, 1, p[i] + 1, min(p[i] + k, n));
		if(t != inf) {
			G[p[i]].push_back(p[t]);
			++in[p[t]];
		}
		t = query(1, n, 1, max(1, p[i] - k + 1), p[i] - 1);
		if(t != inf) {
			G[p[i]].push_back(p[t]);
			++in[p[t]];
		}
		update(1, n, 1, p[i], i);
	}
	priority_queue<int, vector<int>, greater<int> > q;
	for(int i = 1; i <= n; ++i) {
		if(!in[i]) {
			q.push(i);
		}
	}
	while(!q.empty()) {
		int u = q.top();
		ans[u] = ++ans[0];
		q.pop();
		for(int i = 0; i < G[u].size(); ++i) {
			int v = G[u][i];
			if(!--in[v]) {
				q.push(v);
			}
		}
	}
	for(int i = 1; i <= n; ++i) {
		printf("%d\n", ans[i]);
	}
	return 0;
}