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 B {

	MyScanner in;
	PrintWriter out;
	final static String filename = "";

	class MyScanner {
		BufferedReader br;
		StringTokenizer st;

		public MyScanner(String file) {
			try {
				br = new BufferedReader(new FileReader(file));
				st = new StringTokenizer("");
			} catch (FileNotFoundException e) {
				e.printStackTrace();
				System.exit(1);
			}
		}

		public MyScanner(InputStream is) {
			br = new BufferedReader(new InputStreamReader(is));
			st = new StringTokenizer("");
		}

		String nextToken() {
			try {
				while (st == null || !st.hasMoreTokens())
					st = new StringTokenizer(br.readLine());

			} catch (IOException e) {
				e.printStackTrace();
				System.exit(1);
			}

			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(this.nextToken());
		}

		double nextDouble() {
			return Double.parseDouble(this.nextToken());
		}

		long nextLong() {
			return Long.parseLong(this.nextToken());
		}

		void close() throws IOException {
			br.close();
		}
	}

	int[] a;
	int n;

	void solve() throws IOException {
		n = in.nextInt();
		int k = in.nextInt();
		int[] b = new int[n];
		a = new int[n];
		for (int i = 0; i < n; i++)
			b[i] = in.nextInt();
		boolean[] used = new boolean[n];
		for (int i = 0; i < n; i++) {
			int t = -1;
			for (int j = 0; j < n; j++)
				if (b[j] == 0 && !used[j]) {
					t = j;
					break;
				}
			used[t] = true;
			a[i] = t;
			for (int j = 0; j < n; j++)
				if (j + k <= t)
					b[j]--;
		}
		for (int i = 0; i < n; i++)
			out.print((a[i] + 1) + " ");
	}

	void run() {
		try {
			in = new MyScanner(System.in);
			out = new PrintWriter(System.out);

			solve();

			in.close();
			out.close();
		} catch (IOException e) {
			e.printStackTrace();
			System.exit(1);
		}

	}

	public static void main(String[] args) {
		new B().run();
	}

}
