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 <cstdio>
#include <cstring>

#include <algorithm>
#include <vector>
#include <queue>

#define PB push_back

using namespace std;

const int N = 5e5 + 500;
const int OFF = (1 << 19);

int n, k, P[N], Q[N], T[2 * OFF], sol[N];

inline int miin(int x, int y){
	if(Q[x] < Q[y])
		return x;
	return y;
}

void update(int i,int x){
	Q[i] = x; T[OFF + i] = i;
	for(i = (i + OFF) / 2; i ; i /= 2)
		T[i] = miin(T[2 * i], T[2 * i + 1]);
}

int query(int i,int a,int b,int lo,int hi){
	if(lo <= a && b <= hi) return T[i];
	if(a > hi || b < lo) return n;
	return miin(query(2 * i, a, (a + b) / 2, lo, hi), query(2 * i + 1, (a + b) / 2 + 1, b, lo, hi));
}

inline int lokal(int x){
	return x == query(1, 0, OFF - 1, max(0, x - k + 1), min(n - 1, x + k - 1));
}

priority_queue < pair < int, int > , vector < pair < int, int > >, less < pair < int, int > > > PQ;

inline void buket(int x){
	if(x < 0 || x * k >= n)
		return;
	int kan = query(1, 0, OFF - 1, x * k, min(x * k + k - 1, n - 1));
	if(Q[kan] == n)
		return;
	if(lokal(kan))
		PQ.push({kan, kan});
}

int main(){
	scanf("%d%d", &n, &k);
	for(int i = 0;i < n;i++){
		scanf("%d", P + i); P[i]--;
		update(i, n - P[i] - 1);
	}
	update(n, n);
	for(int i = 0;i * k < n;i++)
		buket(i);
	int trn = n - 1;
	for(;!PQ.empty();){
		int cur = PQ.top().second; PQ.pop();
		if(Q[cur] == n)
			continue;
		sol[cur] = trn--;
		update(cur, n);
		buket(cur / k);
		buket(cur / k - 1);
		buket(cur / k + 1);
	}
	for(int i = 0;i < n;i++)
		printf("%d\n", sol[i] + 1);
}




