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 Solution {
	
	private BufferedReader in;
	private PrintWriter out;
	private StringTokenizer st;

	void solve() throws IOException {
		int n = nextInt();
		int k = nextInt();
		int[] b = new int[n];
		for (int i = 0; i < n; ++i) {
			b[i] = nextInt();
		}
		ArrayList<Integer> a = new ArrayList<Integer>();
		for (int i = n - 1; i >= 0; --i) {
			for (ListIterator<Integer> it = a.listIterator(); ;) {
				if (b[i] == 0) {
					it.add(i);
					break;
				}
				int x = it.next();
				if (x >= i + k) {
					b[i]--;
				}
			}
		}
		for (int i : a) {
			out.print((i + 1) + " ");
		}
		out.println();
	}

	Solution() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		
		eat("");
		
		solve();
		
		in.close();
		out.close();
	}
	
	private void eat(String str) {
		st = new StringTokenizer(str);
	}
	
	String next() throws IOException {
		while (!st.hasMoreTokens()) {
			String line = in.readLine();
			if (line == null) {
				return null;
			}
			eat(line);
		}
		return st.nextToken();
	}
	
	int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
	
	long nextLong() throws IOException {
		return Long.parseLong(next());
	}
	
	double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

	public static void main(String[] args) throws IOException {
		new Solution();
	}
}
