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 a[1010], b[1010];
int ans[1010], ans2[1010];
bool temp[1010];
int s;
int main() {
  cin >> n >> k;
  s = 0;
  for (int i = 1; i <= n; i++) ans[i] = i;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) b[i] = i + k;
  for (int i = n - k; i > 0; i--) {
    if (a[i]) {
      int t = 0, j;
      int p = ans[i];
      for (j = i + 1; j <= n; j++) {
        ans[j - 1] = ans[j];
        if (ans[j] >= b[i]) {
          t++;
          if (a[i] == t) break;
        }
      }
      ans[j] = p;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}
