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.*;
import java.io.*;
import java.math.*;

public class Main implements Runnable {
	final String filename = "test";

	void solve() throws Exception {
		n = iread();
		k = iread();
		a = new int[n];
		b = new int[n];
		for (int i = 0; i < n; i++)
			b[i] = iread();
		int[] c = new int[n + k];
		was = new boolean[n];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				if (was[j])
					continue;
				if (c[j] == b[j]) {
					a[i] = j;
					was[j] = true;
					for (int s = 0; s <= j - k; s++)
						c[s]++;
					break;
				}
			}
		}
		for (int i = 0; i < n; i++)
			out.write((a[i] + 1) + " ");
		out.write("\n");
	}

	int n, k;
	int[] b;
	int[] a;
	boolean[] was;

	public static void main(String[] args) {
		Main inst = new Main();
		new Thread(inst).run();
	}

	int iread() {
		return Integer.parseInt(readword());
	}

	long lread() {
		return Long.parseLong(readword());
	}

	double dread() {
		return Double.parseDouble(readword());
	}

	String readword() {
		try {
			int c = in.read();
			while (c <= 32) {
				if (c < 0)
					return "";
				c = in.read();
			}
			StringBuilder sb = new StringBuilder();
			while (c > 32) {
				sb.append((char) c);
				c = in.read();
			}
			return sb.toString();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
			return "";
		}
	}

	BufferedReader in;
	BufferedWriter out;

	public void run() {
		try {
			// in = new BufferedReader(new FileReader(filename+".in"));
			// out = new BufferedWriter(new FileWriter(filename+".out"));
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new BufferedWriter(new OutputStreamWriter(System.out));

			solve();
			out.flush();
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

}
