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;
const long maxn = 1005;
long a[maxn];
long b[maxn];
long n, q, choice, k;
bool used[maxn];
int main() {
  scanf("%ld%ld", &n, &k);
  for (q = 1; q <= n; q++) {
    scanf("%ld", &b[q]);
    used[q] = false;
  }
  for (long step = 1; step <= n; step++) {
    for (choice = 1; choice <= n; choice++)
      if (!used[choice]) {
        long temp = k + choice;
        long cht = 0;
        for (long pred = step - 1; pred >= 1; pred--)
          if (a[pred] >= temp) cht++;
        if (cht == b[choice]) break;
      }
    a[step] = choice;
    used[choice] = true;
    printf("%ld ", a[step]);
  }
}
