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.io.*;
import java.math.*;
import java.util.*;
import java.text.*;
public class Main 
{
    static int maxn = 1010;
    static int [] L = new int [maxn];
    static int [] R = new int [maxn];
    static int [] b = new int [maxn];
    public static void main(String[] args) 
    {
        Scanner cin = new Scanner (new BufferedInputStream(System.in));
        int n = cin.nextInt();
        int K = cin.nextInt();
        for(int i=1;i<=n;i++)
            b[i] = cin.nextInt();
        R[0]=n;
        L[n]=0;
        R[n]=n+1;
        L[n+1]=n;
        for(int i=n-1;i>=1;i--)
        {
            if(b[i] == 0)
            {
                L[i]=0;
                R[i]=R[0];
                L[R[i]]=i;
                R[0]=i;
            }
            else
            {
                int total=0;
                for(int k=R[0];k!=n+1;k=R[k])
                {
                    if(k >= i+K)
                        total++;
                    if(total == b[i])
                    {
                        L[i]=k;
                        R[i]=R[k];
                        R[k]=i;
                        L[R[i]]=i;
                        break;
                    }
                }
            }
        }
        for(int k=R[0];k!=n+1;k=R[k])
            System.out.printf(k+" ");
    }
}