Let A = {a1, a2, ..., an} be any permutation of the first n natural numbers {1, 2, ..., n}. You are given a positive integer k and another sequence B = {b1, b2, ..., bn}, where bi is the number of elements aj in A to the left of the element at = i such that aj ≥ (i + k).

For example, if n = 5, a possible A is {5, 1, 4, 2, 3}. For k = 2, B is given by {1, 2, 1, 0, 0}. But if k = 3, then B = {1, 1, 0, 0, 0}.

For two sequences X = {x1, x2, ..., xn} and Y = {y1, y2, ..., yn}, let i-th elements be the first elements such that xi ≠ yi. If xi < yi, then X is lexicographically smaller than Y, while if xi > yi, then X is lexicographically greater than Y.

Given n, k and B, you need to determine the lexicographically smallest A.

Input

The first line contains two space separated integers n and k (1 ≤ n ≤ 1000, 1 ≤ k ≤ n). On the second line are n integers specifying the values of B = {b1, b2, ..., bn}.

Output

Print on a single line n integers of A = {a1, a2, ..., an} such that A is lexicographically minimal. It is guaranteed that the solution exists.

Examples

Input

5 2
1 2 1 0 0


Output

4 1 5 2 3 

Input

4 2
1 0 0 0


Output

2 3 1 4 


import java.io.*;
import java.util.*;
import static java.lang.Math.*;

public class ProblemB_Manthan {
        
    final boolean ONLINE_JUDGE=System.getProperty("ONLINE_JUDGE")!=null;
    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok=new StringTokenizer("");
    
    void init() throws FileNotFoundException{
            if (ONLINE_JUDGE){
                    in=new BufferedReader(new InputStreamReader(System.in));
                    out =new PrintWriter(System.out);
            }
            else{
                    in = new BufferedReader(new FileReader("input.txt"));
                    out = new PrintWriter("output.txt");
            }
    }
    
    String readString() throws IOException{
            while(!tok.hasMoreTokens()){
                    tok=new StringTokenizer(in.readLine());
            }
            return tok.nextToken();
    }
    
    int readInt() throws IOException{
            return Integer.parseInt(readString());
    }
    
    public static void main(String[] args){
            new ProblemB_Manthan().run();
    }
    
    public void run(){
            try{
                    long t1=System.currentTimeMillis();
                    init();
                    solve();
                    out.close();
                    long t2=System.currentTimeMillis();
                    System.err.println("Time = "+(t2-t1));
            }catch (Exception e){
                    e.printStackTrace(System.err);
                    System.exit(-1);
            }
    }
    
    class Number implements Comparable<Number>{
        
        int i;
        int ind;
        
        public Number(int j, int ind){
            i=j;
            this.ind=ind;
        }

        @Override
        public int compareTo(Number o) {
            return ind-o.ind;
        }
    }
    
    void solve() throws IOException{
        int n=readInt();
        int k=readInt();
        ArrayList<Integer> l=new ArrayList<Integer>();
        Number[] num=new Number[n];
        for (int i=0; i<n; i++){
            l.add(i);
            num[i]=new Number(i, i);
        }
        int[] b=new int[n];
        for (int i=0; i<n; i++){
            b[i]=readInt();
        }
        Number[] q;
        for (int i=n-k-1; i>=0; i--){
            q=new Number[n-i-k];
            for (int j=i+k; j<n; j++){
                q[j-i-k]=num[j];
            }
            Arrays.sort(q);
            int ind=-1;
            if (b[i]==0){
                ind=i;
            }else{
                ind=q[b[i]-1].ind;
                for (int j=i+1; j<=ind; j++){
                    num[l.get(j)].ind--;
                }
            }
            l.add(ind+1, i);
            l.remove(i);
            num[i].ind=ind;
        }
        for (int i=0; i<n; i++){
            out.print((l.get(i)+1)+" ");
        }
    }               
}

