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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
int a, b, c, d, i, j, n, m, k;
int mas[1002], ans[1002];
bool used[1002];
int fen[1002];
inline void add(int first, int a) {
  for (int i = first; i <= n; i |= i + 1) fen[i] += a;
}
inline int sum(int first) {
  int ans = 0;
  for (; first >= 0; first = (first & (first + 1)) - 1) ans += fen[first];
  return ans;
}
inline int sum(int l, int r) {
  if (l > r) return 0;
  if (!l)
    return sum(r);
  else
    return sum(r) - sum(l - 1);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int _n((n)-1), i(0); i <= _n; i++) scanf("%d", &mas[i]);
  for (int _n((n)-1), i(0); i <= _n; i++) {
    for (int _n((n)-1), j(0); j <= _n; j++) {
      if (used[j] || sum(j + k, n - 1) != mas[j]) continue;
      used[j] = 1;
      ans[i] = j;
      add(j, 1);
      break;
    }
    if (i) printf(" ");
    printf("%d", ans[i] + 1);
  }
  printf("\n");
}
