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 a[1100], b[1100];
int main() {
  int n, k, d, i, tmp, cnt, j;
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; i++) scanf("%d", b + i);
  for (i = 0; i < k; i++) a[i] = n - k + i + 1;
  int kol = k;
  for (cnt = n - k; cnt > 0; cnt--) {
    for (j = 0, tmp = 0; tmp < b[cnt]; j++)
      if (a[j] >= cnt + k) tmp++;
    for (i = kol; i >= j; i--) a[i] = a[i - 1];
    a[j] = cnt;
    kol++;
  }
  for (i = 0; i < n; i++) printf("%d ", a[i]);
  return 0;
}
