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>
int findmin(int *b, int n, int k);
int main() {
  int i, j, k, n, *b, *a;
  scanf("%d%d", &n, &k);
  b = (int *)malloc(sizeof(int) * n);
  a = (int *)malloc(sizeof(int) * n);
  for (i = 0; i < n; i++) scanf("%d", &b[i]);
  for (i = 0; i < n; i++) {
    j = findmin(b, n, k);
    a[i] = j + 1;
    b[j] = -1;
  }
  for (i = 0; i < n - 1; i++) printf("%d ", a[i]);
  printf("%d\n", a[i]);
  return (0);
}
int findmin(int *b, int n, int k) {
  int i, index = -1;
  for (i = 0; i < n; i++) {
    if (b[i] == 0) {
      index = i;
      break;
    }
  }
  for (i = 0; i < n; i++) {
    if (b[i] > 0 && index >= i + k) b[i]--;
  }
  return (index);
}
