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
//Be Name Khoda
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
template <typename T> using ordered_set =  tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T> using ordered_multiset =  tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
 
typedef long long int LL;
typedef LL ll;
typedef long double ld;
typedef string str;
typedef pair<ll,ll> pll;
typedef vector<ll> vll;
typedef stringstream strs;
 
#define X first
#define Y second
#define PB push_back
#define For(i,a,b) for (int i=a;i<b;i++)
#define Ford(i,a,b) for (int i=a;i>=b;i--)
#define smax(a,b) a=max(a,b)
#define smin(a,b) a=min(a,b)
#define SZ(a) ((ll)a.size())
#define ER(a) cout << #a << ' ' << a << endl
#define LB(a,n,x) (lower_bound(a,(a)+(n),x)-(a))
#define RET(x) { cout << x; exit(0); } 
const ll M=1e6+5,LG=17,SM=600+5,inf=1e18+5;
ll mod=1e9+7;
 
ll n;
ll p[M];
ll ip[M],rot[M],k;

int main()
{
    cin >> n >> k;
    For(i,0,n)
    {
        cin >> p[i];
        p[i]--;
        ip[p[i]]=i;    
    }
    ordered_set<ll> se;
    For(i,0,n)
    {
        rot[i]=se.order_of_key(ip[i]+k);
        rot[i]-=se.order_of_key(ip[i]);
        se.insert(ip[i]);
    }
    For(ii,0,n)
    {
        ll i=p[ii];
        rot[i]=*se.find_by_order(rot[i]);
        se.erase(rot[i]);
        ip[rot[i]]=ii;
    }
    For(i,0,n) cout << ip[i]+1 << '\n';
}    
