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.util.*;

/**
 * Codeforces #67B "Restoration of the Permutation"
 * @link http://codeforces.ru/problemset/problem/67/B
 * @author Caesar
 */
public class Solution67B implements Runnable {

    private static final int EXIT_CODE = -1;

    private StringTokenizer tok;
    private BufferedReader in;
    private PrintWriter out;

    public static void main(String[] args) {
        new Thread(new Solution67B()).start();
    }

    public void run() {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        long start = System.currentTimeMillis();
        try {
            solve();
        } catch (Throwable ex) {
            ex.printStackTrace();
            System.exit(EXIT_CODE);
        } finally {
            out.flush();
        }
        System.err.println("Time used: " + (System.currentTimeMillis() - start));
    }

    private String readString() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    private int readInt() throws IOException {
        return Integer.parseInt(readString());
    }

    public void solve() throws IOException{
        int n = readInt();
        int k = readInt();
        int[] b = new int[n];
        for (int i = 0; i < n; ++i) {
            b[i] = readInt();
        }
        int[] a = new int[n];
        boolean[] used = new boolean[n];
        int[] left = new int[n];
        for (int i = 0; i < n; ++i) {
            int count = 0;
            for (int j = 0; j < i; ++j) {
                if (a[j] >= k) {
                    ++count;
                }
            }
            for (int j = 0; j < n; ++j) {
                int elem = k + j - 1;
                left[j] = elem < n && used[k + j - 1] ? --count : count;
            }
            for (int j = 0; j < n; ++j) {
                if (!used[j] && b[j] == left[j]) {
                    used[j] = true;
                    a[i] = j + 1;
                    break;
                }
            }
            out.print(a[i]);
            if (i == n) {
                out.println();
            } else {
                out.print(' ');
            }
        }
    }

}
