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 a[1004], b[1004];
int main() {
  int i, n, k, m, j;
  scanf("%d %d", &n, &k);
  for (i = (0); i < (n); i++) scanf("%d", &a[i]);
  for (i = (n - k + 1); i <= (n); i++) b[i - (n - k + 1)] = i;
  for (i = (n - k); i >= (1); i--) {
    j = 0;
    m = a[i - 1];
    while (m) {
      if (b[j] >= i + k) m--;
      j++;
    }
    memmove(&b[j + 1], &b[j], (n - i - j) * sizeof(int));
    b[j] = i;
  }
  for (i = (0); i < (n - 1); i++) printf("%d ", b[i]);
  printf("%d\n", b[n - 1]);
  return 0;
}
