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() {
  vector<int> arr, ord;
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    int t;
    cin >> t;
    arr.push_back(t);
  }
  for (int i = n - 1; i >= 0; --i) {
    int need = arr[i], cnt = 0;
    for (int j = 0; j <= ord.size(); ++j) {
      if (need == cnt) {
        ord.insert(ord.begin() + j, i);
        break;
      }
      cnt += (ord[j] >= i + m);
    }
  }
  for (int i = 0; i < ord.size(); ++i) {
    cout << ord[i] + 1 << ' ';
  }
}
