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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.ArrayList;
import java.util.List;

public class B61 {

    static StreamTokenizer in;
    static PrintWriter out;
    
    static int nextInt() throws IOException {
        in.nextToken();
        return (int)in.nval;
    }
    
    static String nextString() throws IOException {
        in.nextToken();
        return in.sval;
    }

    public static void main(String[] args) throws IOException {
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);

        int n = nextInt(), k = nextInt();
        int[] b = new int[n];
        for (int i = 0; i < n; i++)
            b[i] = nextInt();
        
        List<Integer> ans = new ArrayList<Integer>();
        for (int e = n; e >= 1; e--) {
            int q = b[e-1], p = 0;
            while (q > 0)
                if (ans.get(p++) >= e+k)
                    q--;
            ans.add(p, e);
        }
        
        for (int e : ans)
            out.print(e + " ");
        out.println();
        
        out.flush();
    }

}
