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 
/**
 * Created by IntelliJ IDEA.
 * User: Taras_Brzezinsky
 * Date: 9/17/11
 * Time: 10:53 AM
 * To change this template use File | Settings | File Templates.
 */

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.StringTokenizer;

public class TaskB extends Thread {
    public TaskB() {
        this.input = new BufferedReader(new InputStreamReader(System.in));
        this.output = new PrintWriter(System.out);
        this.setPriority(Thread.MAX_PRIORITY);
    }

    private void solve() throws Throwable {
        int n = nextInt();
        int[] has = new int[n];
        boolean[] was = new boolean[n];
        int k = nextInt();
        for (int i = 0; i < n; ++i) {
            has[i] = nextInt();
        }
        for (int served = 0; served < n; ++served) {
            for (int i = 0; i < n; ) {
                if (has[i] == 0 && !was[i]) {
                    output.print((i + 1) + " ");
                    was[i] = true;
                    for (int affected = i - k; affected >= 0; --affected) {
                        --has[affected];
                    }
                    i = n;
                } else {
                    ++i;
                }
            }
        }
    }

    public void run() {
        try {
            solve();
        } catch (Throwable e) {
            System.err.println(e.getMessage());
            e.printStackTrace();
            System.exit(666);
        } finally {
            output.flush();
            output.close();
        }
    }

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

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

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

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

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

    private BufferedReader input;
    private PrintWriter output;
    private StringTokenizer tokens = null;
}
