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 n, k;
  cin >> n >> k;
  vector<int> b(n);
  int maxs = -1;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    maxs = max(b[i], maxs);
  }
  vector<int> res;
  for (int i = 0; i <= maxs; i++) {
    vector<int> can;
    vector<int> tempres;
    vector<int> used(res.size());
    for (int j = 0; j < n; j++)
      if (b[j] == i) can.push_back(j);
    if (can.size() == 0) continue;
    if (res.size() == 0) {
      res = can;
      continue;
    }
    int s = 0;
    int l = i;
    for (int j = 0; j < can.size(); j++) {
      while (s < res.size() && (l)) {
        if (res[s] >= k + can[j]) l--;
        if (used[s] == 0) tempres.push_back(res[s]);
        used[s] = 1;
        s++;
      }
      if (s == res.size()) {
        tempres.push_back(can[j]);
      } else {
        tempres.push_back(can[j]);
        l = i;
        s = 0;
      }
    }
    for (int t = 0; t < res.size(); t++)
      if (used[t] == 0) tempres.push_back(res[t]);
    res = tempres;
  }
  for (int i = 0; i < res.size(); i++) cout << res[i] + 1 << ' ';
  return 0;
}
