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.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;

public class B {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.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[] a = new int[n];
        Arrays.fill(a, 0);
        for (int i = n - 1; i >= 0; --i) {
            int bigger = 0, pos = 0;
            for (; bigger < b[i]; ++pos) {
                if (a[pos] >= i + 1 + k) {
                    ++bigger;
                }
            }
            --pos;
            while (pos + 1 < n && a[pos + 1] < i + 1 + k) {
                ++pos;
            }
            int cur = i + 1;
            while (cur > 0) {
                if (a[pos] < cur || a[pos] >= cur + k) {
                    int temp = a[pos];
                    a[pos] = cur;
                    cur = temp;
                }               
                --pos;
            }
        }
        for (int i = 0; i < n; ++i) {
            out.print(a[i] + " ");
        }
        in.close();
        out.close();
    }   
}
