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.*;
import java.math.*;
public class Main {

	int[] answer(int[] b, int k) {
		if (b.length == 0) {
			return new int[0];
		} else {
			int[] preAns = answer(Arrays.copyOfRange(b, 1, b.length), k);
			int[] ans = new int[b.length];
			int i;
			int c = 0;
			for (i = 0; c < b[0]; ++i) {
				if (preAns[i] >= k) {
					++c;
				}
				ans[i] = preAns[i] + 1;
			}
			ans[i] = 1;
			for (++i; i < ans.length; ++i) {
				ans[i] = preAns[i - 1] + 1;
			}
			return ans;
		}
	}
	void solve() {
		int n = nextInt();
		int k = nextInt();
		int[] b = new int[n];
		for (int i = 0; i < n; ++i) {
			b[i] = nextInt();
		}
		int[] ans = answer(b, k);
		for (int i = 0; i < ans.length; ++i) {
			out.print(ans[i] + " ");
		}
	}
	
	void run() {
		try {
			in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
			out = new PrintWriter(System.out);
		} catch (Exception e){
			e.printStackTrace();
		}
		solve();
		out.close();
	}
	
	StreamTokenizer in;
	PrintWriter out;
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		new Main().run();
	}
	
	int nextInt() {
		try {
			in.nextToken();
		} catch (IOException e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
		}
		return (int)in.nval;
	}
	
	String next() {
		try {
			in.nextToken();
		} catch (IOException e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
		}
		return in.sval;
	}
	
	long nextLong() {
		try {
			in.nextToken();
		} catch (IOException e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
		}
		return (long)in.nval;
	}
	
	double nextDouble() {
		try {
			in.nextToken();
		} catch (IOException e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
		}
		return in.nval;
	}
}
