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 Perestanovka {
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer in;
	static PrintWriter out = new PrintWriter(System.out);
	
	public static String nextToken() throws IOException{
		while (in == null || !in.hasMoreTokens()){
			in = new StringTokenizer(br.readLine());
		}
		
		return in.nextToken();
	}
	
	public static int nextInt() throws IOException{
		return Integer.parseInt(nextToken());
	}
	
	public static double nextDouble() throws IOException{
		return Double.parseDouble(nextToken());
	}
	
	public static void main(String[] args) throws IOException{
		int n = nextInt();
		int k = nextInt();
		
		int[] b = new int[n];
		for (int i = 0; i < b.length; i++) {
			b[i] = nextInt();
		}
		
		ArrayList<Integer> ans = new ArrayList<Integer>();
		
		for (int i = n - k + 1; i <= n; i++) {
			ans.add(i);
		}
		
		up: for (int i = n - k; i > 0; i--) {
			int count = 0;
			for (int z = 0; z < ans.size(); z++) {
				if (count == b[i - 1]){
					ans.add(z, i);
					continue up;
				}
				
				if (ans.get(z) >= i + k){
					count++;
				}
			}
			
			ans.add(i);
		}
		
		for (int u : ans) {
			out.print(u + " ");
		}
		
		out.close();
	}
}
