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.Scanner;

public class Main {
    public static void main(String args[])
    {
        Scanner s=new Scanner(System.in);
        int n=s.nextInt(),k=s.nextInt();
        int[] b=new int[n],a=new int[n];
        for(int i=0;i<n;i++){
             b[i]=s.nextInt();
        }
        for(int i=0;i<n;i++){
               for(int j=0;j<n;j++){
                   if(b[j]==0){
                       a[i]=j+1;
                       b[j]=-1;
                       break;
                   }
                   if(b[j]==-1)continue;
                   int ss=0;
                    for(int l=0;l<i;l++){
                           if(a[l]>=j+k+1)ss++;
                    }
                   if(b[j]==ss)
                   {
                       a[i]=j+1;
                       b[j]=-1;
                       break;
                   }

               }
        }

        for(int i:a)
        {
            System.out.print(i+" ");
        }
        System.out.print(" ");
    }
}