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;
const long long int mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int T = clock();
  long long int n, k;
  cin >> n >> k;
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) cin >> a[i];
  vector<long long int> u(n, 0), ans(n);
  for (long long int i = 0; i < n; i++) {
    long long int j = 0;
    for (j = 0; j < n; j++) {
      if (a[j] == 0 && u[j] == 0) break;
    }
    for (long long int r = 0; r + k <= j; r++) {
      if (u[r] == 0) a[r]--;
    }
    ans[i] = j + 1;
    u[j] = 1;
  }
  for (long long int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cerr << "\n\nTIME: " << (double)(clock() - T) / CLOCKS_PER_SEC << " sec\n";
  T = clock();
  return 0;
}
