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.*;

public class B
{
PrintWriter out;
BufferedReader in;
StringTokenizer ss;

   static void dbg(String s){System.out.println(s);};
   static void deb(long x){System.out.println(x);};
   String next_token() throws IOException
   {while (!ss.hasMoreTokens())ss=new StringTokenizer(in.readLine());
    return ss.nextToken();}

   Double _double() throws IOException
   {return Double.parseDouble (next_token());}

   int _int() throws IOException
   {return Integer.parseInt (next_token());}

   long _long() throws IOException
   {return Long.parseLong (next_token());}

void run()throws IOException
{

in = new BufferedReader(new InputStreamReader(System.in));
out= new PrintWriter(System.out);
String name="";
/*in = new BufferedReader(new FileReader(name+".in"));
out = new PrintWriter(new File(name+".out"));  */
ss = new StringTokenizer(" ");
ss = new StringTokenizer(" ");
int n=_int();
int k=_int();
    int Al[]=new int[n+1];
    int Ar[]=new int[n+1];
    int B[]=new int[n];
    for(int i=0;i<n;i++){
        B[i]=_int();
    }
    int first=n-1;
    Al[n-1]=-1;
    Ar[n-1]=n;
    Al[n]=n-1;

    for(int i=n-2;i>=0;i--){
        int g=first;
        int skoko=0;
        while(skoko<B[i]){
            if(g>=i+k)skoko++;
            g=Ar[g];
        }

        Al[i]=Al[g];
        Al[g]=i;
        Ar[i]=g;
        if(Al[i]==-1)first=i;
        else{
            Ar[Al[i]]=i;
        }
    }
    int G=first;
    for(int i=0;i<n;i++){
        out.print((G+1)+" ");
        G=Ar[G];
    }

     //out.println(Al[3]+" "+Ar[3]);
out.close();
}
public static void main(String[] args)throws Exception
{new B().run();}
}


