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.Locale;
import java.util.StringTokenizer;


public class B implements Runnable {

	BufferedReader in;
	StringTokenizer st = new StringTokenizer("");
	PrintWriter out;

    public 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();
        }
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            int j = 0;
            while (b[j] != 0) {
                j++;
            }
            a[i] = j;
            b[j] = -1;
            for (int t = 0; t <= j - k; t++) {
                b[t]--;
            }
        }
        for (int i = 0; i < n; i++) {
            out.print(a[i]+1 + " ");
        }
    }

    public void run() {
		try {
			Locale.setDefault(Locale.US);
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
			solve();
		} catch (Exception ex) {
			ex.printStackTrace();
			throw new RuntimeException();
		} finally {
			out.close();
		}
	}

	private int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

    private long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    private double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

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

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

}