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.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Locale;
import java.util.StringTokenizer;


public class B implements Runnable {

    
    private void solve() throws IOException {
        int n = nextInt();
        int k = nextInt();
        int[] b = new int[n];
        for (int i = 0; i < n; ++i) {
            b[i] = nextInt();
        }
        ArrayList<Integer> ans = new ArrayList<Integer>();
        for (int i = n - 1; i >= 0; --i) {
            int bal = b[i];
            int j = 0;
            for (j = 0; bal != 0 && j < ans.size(); ++j) {
                if (ans.get(j) >= i + k) {
                    --bal;
                }
            }
            ans.add(j, i);
        }
        for (int i = 0; i < n; ++i) {
            out.print((ans.get(i) + 1) + " ");          
        }
    }
    
    /**
     * @param args
     */
    public static void main(String[] args) {
        new Thread(new B()).start();

    }

    private BufferedReader br;
    private StringTokenizer st;
    private PrintWriter out;

    @Override
    public void run() {
        try {
            Locale.setDefault(Locale.US);
            br = new BufferedReader(new InputStreamReader(System.in));
            st = new StringTokenizer("");
            out = new PrintWriter(System.out);
            solve();
            out.close();            
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(-1);
        }
        
    }

    
    
    String next() throws IOException {
        while (!st.hasMoreTokens()) {
            String temp = br.readLine();
            if (temp == null) {
                return null;
            }
            st = new StringTokenizer(temp);
        }
        return st.nextToken();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
    double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }
    long nextLong() throws IOException {
        return Long.parseLong(next());
    }
    
}
