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 ll long long
#define pb push_back

const int maxn = 5e5 + 20;

int d[maxn] , a[maxn] , n , mx[maxn * 4];

vector<int> adj[maxn];

void update(int p , int val , int s = 0 , int e = n , int v = 1)
{
	if(e - s < 2)
	{
		mx[v] = val;
		return;
	}

	int m = (s + e) / 2;
	if(p < m)
		update(p , val , s , m , 2 * v);
	else
		update(p , val , m , e , 2 * v + 1);

	mx[v] = max(mx[2 * v] , mx[2 * v + 1]);
}

int get(int l , int r , int s = 0 , int e = n , int v = 1)
{
	if(l <= s && e <= r)
		return mx[v];
	if(r <= s || e <= l)
		return -1;

	int m = (s + e) / 2;
	return max(get(l , r , s , m , 2 * v) , get(l , r , m , e , 2 * v + 1));
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	memset(mx , -1 , sizeof mx);

	int k;
	cin >> n >> k;

	for(int i = 0; i < n; i++)
	{
		int x;
		cin >> x;
		x--;

		a[x] = i;
	}

	for(int i = 0; i < n; i++)
	{
		int v = get(a[i] - k + 1 , a[i]);
		if(v >= 0)
			adj[a[v]].pb(a[i]) , d[a[i]]++;

		v = get(a[i] , a[i] + k);
		if(v >= 0)
			adj[a[v]].pb(a[i]) , d[a[i]]++;

		update(a[i] , i);
	}

	set<int> st;
	for(int i = 0; i < n; i++)
		if(!d[i])
			st.insert(i);

	for(int _ = 0; _ < n; _++)
	{
		assert(!st.empty());

		int v = *st.begin();
		st.erase(v);

		for(auto u : adj[v])
		{
			d[u]--;
			if(!d[u])
				st.insert(u);
		}

		a[v] = _;
	}

	for(int i = 0; i < n; i++)
		cout << a[i] + 1 << " ";
	cout << endl;
}














