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.ListIterator;

/**
 * Created by IntelliJ IDEA.
 * User: piyushd
 * Date: 3/13/11
 * Time: 9:43 PM
 * To change this template use File | Settings | File Templates.
 */
public class Task2 {

    void run(){
		int n = nextInt(), k = nextInt();
		int[] b = new int[n];
		for(int i = 0; i < n; i++)
			b[i] = nextInt();

		ArrayList<Integer> res = new ArrayList<Integer>();
		for(int i = n - 1; i >= 0; i--){
			for(ListIterator<Integer> it = res.listIterator(); ;){
				if(b[i] == 0) {
					it.add(i);
					break;
				}
				int x = it.next();
				if(x >= i + k) b[i]--;
			}
		}

		for(int x : res) System.out.print((x + 1) + " ");
		System.out.println();
    }

    int nextInt(){
        try{
            int c = System.in.read();
            if(c == -1) return c;
            while(c != '-' && (c < '0' || '9' < c)){
                c = System.in.read();
                if(c == -1) return c;
            }
            if(c == '-') return -nextInt();
            int res = 0;
            do{
                res *= 10;
                res += c - '0';
                c = System.in.read();
            }while('0' <= c && c <= '9');
            return res;
        }catch(Exception e){
            return -1;
        }
    }

    long nextLong(){
        try{
            int c = System.in.read();
            if(c == -1) return -1;
            while(c != '-' && (c < '0' || '9' < c)){
                c = System.in.read();
                if(c == -1) return -1;
            }
            if(c == '-') return -nextLong();
            long res = 0;
            do{
                res *= 10;
                res += c-'0';
                c = System.in.read();
            }while('0' <= c && c <= '9');
            return res;
        }catch(Exception e){
            return -1;
        }
    }

    double nextDouble(){
        return Double.parseDouble(next());
    }

    String next(){
        try{
            StringBuilder res = new StringBuilder("");
            int c = System.in.read();
            while(Character.isWhitespace(c))
                c = System.in.read();
            do{
                res.append((char)c);
            }while(!Character.isWhitespace(c=System.in.read()));
            return res.toString();
        }catch(Exception e){
            return null;
        }
    }

    String nextLine(){
        try{
            StringBuilder res = new StringBuilder("");
            int c = System.in.read();
            while(c == '\r' || c == '\n')
                c = System.in.read();
            do{
                res.append((char)c);
                c = System.in.read();
            }while(c != '\r' && c != '\n');
            return res.toString();
        }catch(Exception e){
            return null;
        }
    }

    public static void main(String[] args){
        new Task2().run();
    }
}
