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.awt.Point;
import java.awt.geom.Line2D;
import java.awt.geom.Point2D;
import java.awt.Polygon;
import java.io.*;
import java.math.BigInteger;
import static java.math.BigInteger.*;
import java.util.*;
public class B{
	void solve()throws Exception
	{

        int n=nextInt();
        int k=nextInt();
        int[]b=new int[n];
        for(int i=0;i<n;i++)
            b[i]=nextInt();
        int[]res=new int[n];
        boolean []vis=new boolean[n];
        for(int i=0;i<n;i++)
        {
            int cur=-1;
            for(int j=0;j<n;j++)
                if(!vis[j] && b[j]==0)
                {
                    cur=j;
                    break;
                }
            //System.out.println(cur);
            res[i]=cur;

            vis[cur]=true;
            for(int j=0;j<n;j++)
                if(!vis[j] && cur>=j+k)
                {
                    b[j]--;
                }
        }
        for(int i=0;i<n;i++)
        {
            res[i]++;
            if(i==n-1)
                System.out.println(res[i]);
            else
                System.out.print(res[i]+" ");

        }
	}

    BufferedReader reader;
	PrintWriter writer;
	StringTokenizer stk;
	void run()throws Exception
	{
		reader=new BufferedReader(new InputStreamReader(System.in));
		stk=null;
		writer=new PrintWriter(System.out);
		solve();
		reader.close();
		writer.close();
	}
	int nextInt()throws Exception
	{
		return Integer.parseInt(nextToken());
	}
	long nextLong()throws Exception
	{	
		return Long.parseLong(nextToken());
	}
	double nextDouble()throws Exception
	{		
		return Double.parseDouble(nextToken());
	}
	String nextString()throws Exception
	{
		return nextToken();
	}
	String nextLine()throws Exception
	{
		return reader.readLine();
	}
	String nextToken()throws Exception
	{
		if(stk==null || !stk.hasMoreTokens())
		{
			stk=new StringTokenizer(nextLine());
			return nextToken();
		}
		return stk.nextToken();
	}
	
	public static void main(String[]a) throws Exception
	{	
		new B().run();

	}
	
}