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 
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF67B extends PrintWriter {
	CF67B() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF67B o = new CF67B(); o.main(); o.flush();
	}

	void main() {
		int n = sc.nextInt();
		int k = sc.nextInt();
		int[] bb = new int[n];
		for (int i = 0; i < n; i++)
			bb[i] = sc.nextInt();
		int[] aa = new int[n];
		int m = 0;
		for (int a = n - 1; a >= 0; a--) {
			int j = 0;
			while (bb[a] > 0) {
				if (aa[j] >= a + k)
					bb[a]--;
				j++;
			}
			for (int j_ = m++; j_ > j; j_--)
				aa[j_] = aa[j_ - 1];
			aa[j] = a;
		}
		for (int i = 0; i < n; i++)
			print(aa[i] + 1 + " ");
		println();
	}
}
