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 
#include <bits/stdc++.h>
using namespace std;
int n, k;
int b[1024], a[1024], val[1024];
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", b + i);
    val[i] = i + 1;
  }
  int sobra = n;
  for (int it = 0; it < n; it++) {
    for (int i = 0; i < sobra; i++) {
      if (b[i] == 0) {
        a[it] = val[i];
        for (int j = i - 1; j >= 0; j--) {
          if (val[j] <= val[i] - k) {
            b[j]--;
          }
        }
        for (int j = i; j < sobra - 1; j++) {
          b[j] = b[j + 1];
          val[j] = val[j + 1];
        }
        sobra--;
        break;
      }
    }
  }
  printf("%d", a[0]);
  for (int i = 1; i < n; i++) {
    printf(" %d", a[i]);
  }
  printf("\n");
  return 0;
}
