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


public class ProblemB {

	public static void main(String[] args) throws IOException {
		
		BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
		
		StringTokenizer st = new StringTokenizer(stdin.readLine());
		int n = Integer.valueOf(st.nextToken());
		int k = Integer.valueOf(st.nextToken());
		
		int[] b = new int[n];
		st = new StringTokenizer(stdin.readLine());
		for (int i = 0; i < n; i++) {
			b[i] = Integer.valueOf(st.nextToken());
		}
	
		boolean[] used = new boolean[n];
		boolean added = true;
		
		LinkedList<Integer> resultList = new LinkedList<Integer>();
		
		while(added) {
			added = false;
			for(int i = 0; i < n; i++) {
				if(!used[i] && b[i] <= 0) {
					used[i] = true;
					added = true;
					resultList.add(i+1);
					for (int j = i - k; j >= 0; j--) {
						b[j]--;
					}
					break;
				}
			}
		}
		
		for (Integer e : resultList) {
			out.print(e + " ");
		}
		
		out.flush();
		out.close();
	}
}
