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 all;
int sm[1024];
int b[1024], a[1024];
bool us[1024];
int sum(int p) {
  p = min(p, n);
  int i, r = 0;
  for (i = p; i > 0; i = (i & (i - 1))) r = r + sm[i];
  return all - r;
}
void add(int p, int x) {
  p = min(p, n);
  int i;
  for (i = p; i <= n; i = i + (i & (-i))) sm[i] = sm[i] + x;
}
int main() {
  int i, j, l;
  all = 0;
  memset(sm, 0, sizeof(sm));
  memset(us, 0, sizeof(us));
  scanf("%d %d", &n, &k);
  for (i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (us[j]) continue;
      if (b[j] == 0) {
        a[i] = j;
        for (l = 1; l <= j - k; l++) {
          if (b[l] > 0) b[l]--;
        }
        us[j] = 1;
        break;
      }
    }
    if (i < n)
      printf("%d ", a[i]);
    else
      printf("%d\n", a[i]);
  }
  return 0;
}
