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 main(int argc, char **argv) {
  vector<pair<int, int> > S;
  int N, K;
  cin >> N >> K;
  for (int i = (0); i < (N); ++i) {
    int first;
    cin >> first;
    S.push_back(make_pair(first, i + 1));
  }
  sort(S.begin(), S.end());
  vector<int> res;
  vector<int> cnt(N, 0);
  vector<bool> used(N, 0);
  for (int i = (0); i < (S.size()); ++i) {
    int ptr = -1, make_pair = 1 << 29;
    ;
    for (int j = (0); j < (N); ++j)
      if (cnt[S[j].second - 1] >= S[j].first && !used[S[j].second - 1])
        if (make_pair > S[j].second) make_pair = S[j].second, ptr = j;
    int p = S[ptr].first, q = S[ptr].second;
    for (int j = (0); j < (N); ++j)
      if (j + 1 + K <= q) cnt[j]++;
    used[q - 1] = 1;
    res.push_back(q);
  }
  for (int i = (0); i < (N); ++i) printf("%d ", res[i]);
  puts("");
  return 0;
}
