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.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;

public class RestorationOfPermutation {

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String[] params = br.readLine().split("\\s");
		int n = Integer.parseInt(params[0]);
		int k = Integer.parseInt(params[1]);

		B[] bs = new B[n];

		int index = 0;
		for (String s : br.readLine().split("\\s")) {
			bs[index] = new B(index + 1, Integer.parseInt(s));
			index++;
		}
		
		B[] bssort = Arrays.copyOf(bs, n);
		Arrays.sort(bssort);

		ArrayList<Integer> a = new ArrayList<Integer>(n);
		for (B b : bssort) {
			if (b.bi == 0) {
				a.add(b.i);
			} else {
				int insertindex = 0;
				int count = 0;
				for (int j = 0, len = a.size(); j < len; j++) {
					if (a.get(j) >= b.i + k) {
						count++;
					}
					if (count >= b.bi) {
						insertindex = j + 1;
						while(insertindex < a.size() && bs[a.get(insertindex) - 1].bi == b.bi){
							insertindex++;
						}
						break;
					}
				}
				a.add(insertindex, b.i);
			}
		}

		for (int i = 0, len = a.size(); i < len; i++) {
			if (i == 0)
				System.out.printf("%d", a.get(i));
			else
				System.out.printf(" %d", a.get(i));
		}
	}
	
	private static class B implements Comparable<B> {
		public int i;
		public int bi;

		public B(int i, int bi) {
			this.i = i;
			this.bi = bi;
		}

		@Override
		public int compareTo(B o) {
			if (this.bi == o.bi) {
				return i < o.i ? -1 : 1;
			} else {
				return this.bi < o.bi ? -1 : 1;
			}
		}
		
		public String toString(){
			return "[" + i + ", " + bi + " ]";
		}
	}

}
