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.util.Arrays;
import java.util.Scanner;


public class Main {
	public static void main(String[] args) {
		Scanner r = new Scanner(System.in);
		
		int n = r.nextInt(), k = r.nextInt();
		
		int[] b = new int[n];
		for(int i = 0; i < n; i++)
			b[i] = r.nextInt();
		
		int[] a = new int[n];int p = 0;
		boolean[] v = new boolean[n];
		
		int counter = 0;
		while(counter < n){
			for(int i = 1; i <= n; i++)if(!v[i-1]){
				if(b[i-1] == 0){
					a[p++] = i;
					v[i-1] = true;
					counter++;
					
					for(int j = 1; j <= n; j++){
						if(b[j-1] == 0)continue;
						if(a[p-1] >= j+k)b[j-1]--;
					}
					break;
				}
			}
		}
		
		for(int i = 0;i < n; i++){
			System.out.print(a[i]+" ");
		}
	}
}
