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.util.*;
import java.io.*;
import java.math.BigInteger;
import static java.lang.Math.*;

public class Main implements Runnable {
	StringTokenizer tokenizer = new StringTokenizer("");
	BufferedReader in;
	//PrintWriter out;
	//String problem = "kth";
    PrintStream out;
    long m;

	void debug(String s) {
		System.out.println(s);
	}

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




	public void run() {
		try {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintStream(System.out);
            int n = nextInt();
            int k = nextInt();
            int b [] = new int [n];
            for (int i=0; i<n; i++) b[i] = nextInt();
            for (int i=0; i<n; i++){
                int j=0;
                while(b[j]!=0) j++;
                out.print((j+1)+" ");
                for (int t=0; t<=j-k; t++){
                    b[t]--;
                }
                b[j]--;
            }
		} catch (Exception e) {
			e.printStackTrace();
		}
	}

	boolean seekForToken() {
		while (!tokenizer.hasMoreTokens()) {
			String s = null;
			try {
				s = in.readLine();
			} catch (Exception e) {
				e.printStackTrace();
			}
			if (s == null)
				return false;
			tokenizer = new StringTokenizer(s);
		}
		return true;
	}

	String nextToken() {
		return seekForToken() ? tokenizer.nextToken() : null;
	}

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

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

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

	BigInteger nextBig() {
		return new BigInteger(nextToken());
	}
}
