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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskB {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int k = in.nextInt();
            int[] b = new int[n];
            for (int i = 0; i < n; i++) {
                b[i] = in.nextInt();
            }
            int[] ans = new int[n];
            boolean[] used = new boolean[n];
            for (int i = 0; i < n; i++) {
                int next = 0;
                for (; next <= n && (b[next] != 0 || used[next]); next++) ;
                for (int j = 0; j <= next - k; j++) {
                    b[j]--;
                }
                used[next] = true;
                ans[i] = next + 1;
            }
            for (int i = 0; i < n; i++) {
                out.print(ans[i] + " ");
            }
        }

    }

    static class InputReader {
        private BufferedReader reader;
        private StringTokenizer stt;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
        }

        public String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                return null;
            }
        }

        public String next() {
            while (stt == null || !stt.hasMoreTokens()) {
                stt = new StringTokenizer(nextLine());
            }
            return stt.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}

