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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.ArrayList;

public class Main {
	private static StreamTokenizer in = new StreamTokenizer(new BufferedReader(
			new InputStreamReader(System.in)));
	private static PrintWriter out = new PrintWriter(System.out);

	private static int nextInt() throws Exception {
		in.nextToken();
		return (int) in.nval;
	}

	private static String nextString() throws Exception {
		in.nextToken();
		return in.sval;
	}

	public static void main(String[] args) throws Exception {
		int n = nextInt();
		int k = nextInt();
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = nextInt();
		}
		ArrayList<Integer> w = new ArrayList<Integer>();
		for (int i = n - k; i < n; i++) {
			w.add(i+1);
		}
		for (int e = n - k - 1; e >= 0; e--) {
			
			int p = 0, in = 0;
			for (int j : w) {
				if (p == a[e]) {
					break;
				}
				if (j >= (e + k + 1)) {
					p++;
				}
				in++;
				
			}
			w.add(in, e+1);
		}
		for (int j : w) {
			out.print(j + " ");
		}
		out.println();
		out.flush();
	}
}
