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 
//package manthan;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class B {

	static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer st;
	static PrintWriter out = new PrintWriter(System.out);
	
	static String nextToken() throws IOException{
		while (st==null || !st.hasMoreTokens()){
			String s = bf.readLine();
			if (s == null)
				return null;
			st = new StringTokenizer(s);
		}
		
		return st.nextToken();
	}
	
	static int nextInt() throws IOException{
		return Integer.parseInt(nextToken());
	}
	
	static String nextStr() throws IOException{
		return nextToken();
	}
	
	static double nextDouble() throws IOException{
		return Double.parseDouble(nextToken());
	}
	
	public static void main(String[] args) throws IOException{
		int n = nextInt(),
			k = nextInt(),
			b[] = new int[n],
			a[] = new int[n],
			c[] = new int[n],
			d[] = new int[n];
		
		for (int i=0; i<n; i++)
			b[i] = nextInt();
		
		for (int i=0; i<n; i++)
			for (int l=0; l<n; l++){
				if (c[l]==0 && d[l]==b[l]){
					a[i] = l;
					c[l] = 1;
					for (int j=0; j<=l-k; j++)
						d[j]++;
					break;
				}
			}
		
		for (int i=0; i<n; i++){
			out.print(a[i]+1);
			out.print(' ');
		}
		out.flush();
	}

}
