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 

/**
 *
 * @author jtimv
 */
import java.io.*;
import java.util.*;

public class B {

    StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));

    int nextInt() throws IOException {
        in.nextToken();
        return (int) in.nval;
    }

    public static void main(String[] args) throws IOException {
        new B().main();
    }
    boolean[] yes;
    int b[], num[];
    int k, n;


    void main() throws IOException {
        n = nextInt();
        k = nextInt();
        b = new int[n];
        num = new int[n];
        yes = new boolean[n];
        for (int i = 0; i < b.length; i++) {
            b[i] = nextInt();
            num[i] = i + 1;
        }
        LinkedList<Integer> ans = new LinkedList<Integer>();
        while (ans.size() < n){
            for (int i=n-1; i>=0; i--){
                if (!yes[i]){
                    yes[i] = tryInsert(num[i], b[i], ans);
                }
            }
        }
        boolean printed = false;
        for (Integer i : ans) {
            if (printed) {
                System.out.print(' ');
            } else {
                printed = true;
            }
            System.out.print(i);
        }
        System.out.println();
    }

    boolean tryInsert(int x, int bx, List<Integer> L){
        ListIterator it = L.listIterator();
        int t=0;
        while (t<bx){
            if (!it.hasNext()){
                return false;
            }
            if ((Integer)it.next() >= x+k){
                t++;
            }
        }
        it.add(x);
        return true;
    }

}
