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.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {

        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String[] S = in.readLine().split(" ");
        int n = Integer.parseInt(S[0]);
        int k = Integer.parseInt(S[1]);
        int[] Ans = new int[n];
        int[] A = new int[n];
        S = in.readLine().split(" ");
        for (int i = 0; i < n; i++)
            A[i] = Integer.parseInt(S[i]);
        Ans[0] = n;
        int c = 1;
        for (int i = n - 2; i >= 0; i--) {
            if (A[i] == 0) {
                for (int j = c; j > 0; j--)
                    Ans[j] = Ans[j - 1];
                Ans[0] = i + 1;
            } else {
                int temp = k + i + 1;
                int d = 0;
                for (; A[i] > 0; d++)
                    if (Ans[d] >= temp)
                        A[i]--;
                for (int j = c; j > d; j--)
                    Ans[j] = Ans[j - 1];
                Ans[d] = i + 1;
            }
            c++;
        }
        for (int i = 0; i < n; i++)
            System.out.print(Ans[i] + " ");
        System.out.println();
    }
}
