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 = 5e5 + 9;

int n, A[N], k, q[N], h = 1, t, d[N], Gree[N], ans[N], tot, u;
set<int>P;
vector<int>V[N];
priority_queue<int, vector<int>, greater<int>>Q;

void AddEdge (int u, int v) {
	V[u].push_back(v); ++Gree[v];
}

int main () {
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; ++i) scanf("%d", &A[i]), d[A[i]] = i;
	for (int i = 1; i <= n; ++i) {
		if (i - k > 0) P.erase(A[i - k]);
		auto p = P.upper_bound(A[i]);
		if (p != P.end()) AddEdge(i, d[*p]);
		P.insert(A[i]);
	}
	P.clear();
	for (int i = n; i; --i) {
		if (i + k <= n) P.erase(A[i + k]);
		auto p = P.upper_bound(A[i]);
		if (p != P.end()) AddEdge(i, d[*p]);
		P.insert(A[i]);
	}
	for (int i = 1; i <= n; ++i) if (!Gree[i]) Q.push(i);
	while (!Q.empty()) {
		u = Q.top(); Q.pop();
		ans[u] = ++tot;
		for (auto v : V[u]) if (--Gree[v] == 0) Q.push(v);
	}
	for (int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
	return 0;
}