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.*;

/**
 * @author �����
 *
 */
public class B implements Runnable {

	BufferedReader br;
	StringTokenizer st;
	PrintWriter out;

	String nextToken() throws IOException {
		while (!st.hasMoreTokens()) {
			st = new StringTokenizer(br.readLine());
		}
		return st.nextToken();
	}

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

	String nextStr() throws IOException {
		return nextToken();
	}

	char nextChar() throws IOException {
		return nextToken().charAt(0);
	}

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

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

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

	public void run() {
		try {
			br = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(new OutputStreamWriter(System.out));
			st = new StringTokenizer("");
			solve();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		} finally {
			out.close();
		}
	}

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