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.*;

public class B {

	private void solve() throws IOException {
		n = nextInt();
		k = nextInt();
		b = new int[n];
		for (int i = 0; i < n; i++) {
			b[i] = nextInt();
		}
		next = new int[n + 1];
		prev = new int[n + 1];
		start = 0;
		Arrays.fill(next, -1);
		Arrays.fill(prev, -1);
		get(0);
		for (int i = start; i != n; i = next[i]) {
			out.print((i + 1) + " ");
		}
		out.println();
	}

	int n;
	int k;
	int[] next, prev, b;
	int start;

	void get(int off) {
		if (off == n - 1) {
			start = n - 1;
			next[start] = n;
			prev[n] = start;
			return;
		}
		get(off + 1);
		if (b[off] == 0) {
			next[off] = start;
			prev[start] = off;
			start = off;
			return;
		}
		int cnt = 0;
		int i = start;
		while (cnt < b[off]) {
			if (i >= off + k)
				cnt++;
			i = next[i];
		}
		i = prev[i];
		prev[next[i]] = off;
		next[off] = next[i];
		next[i] = off;
		prev[off] = i;
	}

//	void debug() {
//		for (int i = start; i != n; i = next[i]) {
//			System.err.print(i + " ");
//		}
//		System.err.println();
//	}

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

	BufferedReader br;
	StringTokenizer st;
	PrintWriter out;
	boolean eof = false;

	public void run() {
		try {
			br = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			st = new StringTokenizer("");
			solve();
			out.close();
		} catch (Throwable e) {
			e.printStackTrace();
			System.exit(239);
		}
	}

	String nextToken() throws IOException {
		while (!st.hasMoreTokens()) {
			String line = br.readLine();
			if (line == null) {
				eof = true;
				line = "0";
			}
			st = new StringTokenizer(line);
		}
		return st.nextToken();
	}

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

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

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