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.BufferedWriter;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Locale;
import java.util.StringTokenizer;

@SuppressWarnings("unchecked")
public class Main implements Runnable {
	private void solve() throws Exception {
		int n = nextInt();
		int k = nextInt();
		int[] b = new int[n];
		for (int i = 0; i < n; ++i) {
			b[i] = nextInt();
		}
		int[] a = new int[n];
		Arrays.fill(a, -1);
		boolean[] used = new boolean[n];
		int[] more = new int[n];
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < n; ++j) {
				if (used[j] || more[j] != b[j]) {
					continue;
				}
				a[i] = j;
				used[j] = true;
				break;
			}
			for (int j = 0; a[i] >= j + k; ++j) {
				++more[j];
			}
		}
		for (int i = 0; i < n; ++i) {
			print((a[i]+1) + " ");
		}
		println("");
	}

	private BufferedReader reader;
	private PrintWriter writer;
	private StringTokenizer tokenizer;

	public void run() {
		try {
			reader = new BufferedReader(new InputStreamReader(System.in));
//			reader = new BufferedReader(new FileReader("a.in"));
			writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

			solve();

			reader.close();
			writer.close();
		} catch (Throwable e) {
			throw new AssertionError(e);
		}
	}

	private void print(final Object o) {
		writer.print(o);
	}

	private void println(final Object o) {
		writer.println(o);
	}

	private void printf(final String format, final Object... o) {
		writer.printf(format, o);
	}

	private double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	private int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	private long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	private String nextToken() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			tokenizer = new StringTokenizer(reader.readLine());
		}
		return tokenizer.nextToken();
	}

	public static void main(String[] args) throws InterruptedException {
		final long startTime = System.currentTimeMillis();
		Locale.setDefault(Locale.US);
		final Thread thread = new Thread(null, new Main(), "", 1 << 23);
		thread.start();
		thread.join();
		System.err.printf("%.3f\n", (System.currentTimeMillis() - startTime) * 0.001);
	}
}
