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 CF2
{
	public static void main(String[] args)
	{
		Scanner cin = new Scanner(System.in);
		int[] seq = new int[1005];
		int[] b = new int[1005];
		int n = cin.nextInt();
		int k = cin.nextInt();
		for (int i=1;i<=n;i++)
			b[i] = cin.nextInt();
		for (int i=n-k+1;i<=n;i++)
			seq[i-n+k] = i;
		for (int i=n-k;i>=1;i--)
		{
			int count = 0;//��¼����
			int aim = i+k;//Ŀ��
			int ptr = 0;
			while(count<b[i])
			{
				ptr++;
				if (seq[ptr]>=aim) count++;
				}
			for (int j=n-i;j>=ptr+1;j--)
				seq[j+1] = seq[j];
			seq[ptr+1] = i;
			}
		for (int i=1;i<=n-1;i++)
			System.out.print(seq[i]+" ");
		System.out.println(seq[n]);
		}
	}