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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Scanner;

public class B {
    static HashMap<Integer, List<Integer>> adj = new HashMap<Integer, List<Integer>>();
    static PriorityQueue<Num> que = new PriorityQueue<Num>();
    public static void main(String[] args) {
        
        Scanner in = new Scanner(System.in);
        
        int N = in.nextInt(), k = in.nextInt();
        boolean [][] g = new boolean[N+1][N+1];
        boolean visted [] = new boolean[N+1];
        ArrayList<Num> list = new ArrayList<Num>(N);
        ArrayList<Num> sol = new ArrayList<Num>(N);
        for(int a=1; a<=N; a++){
            int b= in.nextInt();
                getEdges(b, a, k, list, N);
        }
        
            Num[] nums = new Num[N];
            
            for(int i=0; i<list.size(); i++)
                nums[i] = list.get(0);
        
        Collections.sort(list);
        
        while(!list.isEmpty()){
            
            Num num =list.remove(0);
            for(int i=0; i< list.size(); i++){
                if(app(list.get(i), num, k)){
                    list.get(i).count--;
                }
            }
            
            sol.add(num);
            visted[num.a]= true;
            Collections.sort(list);
        }
        
        
        int i=0;
        for( i=0; i< sol.size()-1; i++)
            System.out.print(sol.get(i).a+ " ");
        
        System.out.println(sol.get(i).a);

                
    
    }

    private static boolean app(Num num, Num a, int k) {
        
        return  a.a >= num.a + k;
    }

    private static void getEdges(int b, int i, int k, ArrayList<Num> list, int N) {

        
        int pre = i + k;
        int count = 0;
        while(pre++ <= N && b--!=0){
            count++;
        }
        list.add(new Num(i, count));
    }

}

class Num implements Comparable<Num> {

    int a;
    int count;

    public Num(int i, int __count) {
        a = i;
        count = __count;
    }

    @Override
    public int compareTo(Num o) {
        return (count - o.count != 0 ? count - o.count :a - o.a);
    }
    
    
    @Override
    public String toString() {
        // TODO Auto-generated method stub
        return a +" -  " + count;
    }

}