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>
const int INF = (int)(1e9);
const long long INFLL = (long long)(1e18);
const double EPS = 1e-13;
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  vector<int> b(n);
  for (int i = 0; i < n; i++) cin >> b[i];
  vector<int> a;
  for (int i = 0; i < n; i++)
    if (b[i] == 0) a.push_back(i);
  for (int i = n - 1; i >= 0; i--)
    if (b[i] > 0) {
      int cnt = 0;
      int index = -1;
      for (int j = 0; j < a.size(); j++) {
        if (a[j] >= i + k) cnt++;
        if (cnt == b[i]) {
          index = j + 1;
          break;
        }
      }
      a.insert(a.begin() + index, i);
    }
  for (int i = 0; i < n; i++) cout << a[i] + 1 << " ";
  cout << endl;
  return 0;
}
