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.*;
import java.awt.*;

public class B {

    MyScanner in;
    PrintWriter out;

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

    public void run() throws Exception {
        in = new MyScanner();
        out = new PrintWriter(System.out);

        solve();

        out.close();
    }

    public void solve() throws Exception {
        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[] a = new int[n];
        for (int i = n - 1; i >= 0; i--) {
            int t = b[i];
            int more = 0;
            int j = 0;
            while (more < t) {
                if (a[j] >= i + k) {
                    more++;
                }
                j++;
            }
            for (int l = n - 1; l > j; l--) {
                a[l] = a[l - 1];
            }
            a[j] = i;
        }
        
        for (int i = 0; i < n; i++) {
            out.print(a[i] + 1 + " ");
        }
    }

    class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() throws Exception {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() throws Exception {
            if ((st == null) || (!st.hasMoreTokens())) {
                st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
        }

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

        double nextDouble() throws Exception {
            return Double.parseDouble(next());
        }

        boolean nextBoolean() throws Exception {
            return Boolean.parseBoolean(next());
        }

        long nextLong() throws Exception {
            return Long.parseLong(next());
        }
    }

}
