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>
#pragma GCC optimize ("O2")
#pragma GCC optimize ("unroll-loops")
//#pragma GCC optimize("no-stack-protector,fast-math")

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef pair<ll, ll> pll;
#define debug(x) cerr<<#x<<'='<<(x)<<endl;
#define debugp(x) cerr<<#x<<"= {"<<(x.first)<<", "<<(x.second)<<"}"<<endl;
#define debug2(x, y) cerr<<"{"<<#x<<", "<<#y<<"} = {"<<(x)<<", "<<(y)<<"}"<<endl;
#define debugv(v) cerr<<#v<<" : ";for (auto x:v) cerr<<x<<' ';cerr<<endl;
#define all(x) x.begin(), x.end()
#define pb push_back
#define kill(x) return cout<<x<<'\n', 0;

const ld eps=1e-7;
const int inf=1000000010;
const ll INF=10000000000000010LL;
const int mod = 1000000007;
const int MAXN = 500010, LOG=20;

int n, m, k, u, v, x, y, t, a, b;
int P[MAXN], ans[MAXN];
int deg[MAXN];
vector<int> G[MAXN];
set<pii> st;
priority_queue<int, vector<int>, greater<int>> pq;

void add_edge(int u, int v){
	if (!u || v==n+1) return ;
	G[u].pb(v);
	deg[v]++;
	//debug2(u, v)
}

int main(){
	ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	cin>>n>>k;
	st.insert({0, 0});
	st.insert({n+1, n+1});
	for (int i=1; i<=n; i++){
		cin>>P[i];
		if (i>k) st.erase({P[i-k], i-k});
		auto it=st.lower_bound({P[i], 0});
		add_edge(i, it->second);
		it--;
		add_edge(it->second, i);
		st.insert({P[i], i});
	}
	
	for (int i=1; i<=n; i++) if (!deg[i]) pq.push(i);
	/*while (pq.size())*/for (int i=1; i<=n; i++){
		int v=pq.top();
		pq.pop();
		ans[v]=i;
		for (int u:G[v]) if (!--deg[u]) pq.push(u);
	}
	for (int i=1; i<=n; i++) cout<<ans[i]<<'\n';
	
	return 0;
}
