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>
#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
using namespace std;
typedef long long ll; 
typedef long double ld;
typedef pair<ll, ll> pll;

const ll N = 5e5 + 30, Mod = 1e9 + 7;
const ll SQ = 330;

ll n, k;
ll d[N];
ll a[N], q[N];
set<pll> s;
vector<ll> G[N];
void Add(ll v, ll u) {
	if (!~v || u > n) 
		return;
	G[v].pb(u);
	d[u] ++;
}
int main () {
    ios::sync_with_stdio(0), cin.tie(0);
   	cin >> n >> k;
   	s.insert({-1, -1});
   	s.insert({n + n, n + n});
   	for (int i = 1; i <= n; i ++) {
   		cin >> a[i];
		if (i > k)
			s.erase({a[i - k], i - k}); 
		auto it = *s.lower_bound({a[i], -1});
		ll x = it.S;
		Add(i, x);
		auto itt = s.lower_bound({a[i], -1});
		itt--;
		x = itt->S;
		Add(x, i);
		s.insert({a[i], i});
	}
	set<ll> Q;
	for (int i = 1; i <= n; i ++)
		if (!d[i])
			Q.insert(i);
	for (int i = 1; i <= n; i ++) {
		ll x = *Q.begin();
		Q.erase(x);
		q[x] = i;
		for (auto u : G[x]) {
			d[u] --;
			if (!d[u])
				Q.insert(u);
		}
	}
	for (int i = 1; i <= n; i ++)
		cout << q[i] << '\n';
    return (0);
}