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 = 500010;

int n, p[maxn], ip[maxn], mx[maxn << 2], res[maxn], k, _;
int deg[maxn], l[maxn], e;

priority_queue<int> pq;

struct Edge {
	int v, x;
} E[maxn << 1];

void update(int p, int v, int l, int r, int rt) {
	mx[rt] = max(mx[rt], v);
	if (l == r) return;
	int m = (l + r) >> 1;
	if (p <= m) update(p, v, l, m, rt<<1);
	else update(p, v, m+1, r, rt<<1|1);
	mx[rt] = max(mx[rt<<1], mx[rt<<1|1]);
}

int query(int L, int R, int l, int r, int rt) {
	if (L <= l && r <= R) return mx[rt];
	int m = (l + r) >> 1;
	int ret = 0;
	if (L <= m) ret = max(ret, query(L, R, l, m, rt<<1));
	if (R > m) ret = max(ret, query(L, R, m+1, r, rt<<1|1));
	return ret;
}

inline void addEdge(int u, int v) {
	E[e].v = v, E[e].x = l[u], l[u] = e++;
}

int main() {
	memset(l, -1, sizeof(l));
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &p[i]);
		ip[p[i]] = i;
	}
	for (int i = 1; i <= n; i++) {
		{
			int x = query(max(1, ip[i] - k + 1), ip[i], 1, n, 1);
			if (x) {
				addEdge(ip[i], ip[x]);
				++ deg[ip[x]];
			}
		}
		{
			int x = query(ip[i], min(n, ip[i] + k - 1), 1, n, 1);
			if (x) {
				addEdge(ip[i], ip[x]);
				++ deg[ip[x]];
			}
		}
		update(ip[i], i, 1, n, 1);
	}
	for (int i = 1; i <= n; i++) if (!deg[i]) pq.push(i);
	while (!pq.empty()) {
		int u = pq.top(); pq.pop();
		res[u] = n - (_ ++);
		for (int p = l[u]; p >= 0; p = E[p].x) {
			int v = E[p].v;
			-- deg[v];
			if (!deg[v]) pq.push(v);
		}
	}
	for (int i = 1; i <= n; i++) printf("%d\n", res[i]);
	return 0;
}