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.lang.Math;
import java.util.Scanner;

public class main
{
	public static Scanner in;
	public static PrintStream out;
	
	public static void test()
	{
		int n = in.nextInt();
		int k = in.nextInt();
		int[] B = new int[n];
		int i,j;
		for (i=0; i<n; i++)
		{
			B[i] = in.nextInt();
		}
		
		int[] R = new int[n];
		
		for (i=0; i<k; i++)
		{
			R[i] = n-i;
		}
		
		int p = k;
		int c;
		for (i=n-k; i>0; i--)
		{
			c = B[i-1];
			j = p;
			
			while (c>0)
			{
				R[j] = R[j-1];
				if (i+k <= R[j])
				{
					c--;
				}
				
				j--;
			}
			
			R[j] = i;
			p++;
		}
		
		for (i=n; i>0; i--)
		{
			if (i<n)
			{
				out.print(' ');
			}
			
			out.print(R[i-1]);
		}
		out.println();
		
	}
	
	public static void main(String args[])
	{
		try
		{
			//in = new Scanner(new File("in.txt"));
			//out = new PrintStream(new File("out.txt"));			
			in = new Scanner(System.in);
			out = System.out;
		}
		catch (Exception e)
		{
			return;
		}
		
		
		test();
	}
}
