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() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int n, p;
  cin >> n >> p;
  vector<int> v(n + 1);
  for (int i = 1; i <= n; i++) cin >> v[i];
  vector<set<int> > vs(n + 2);
  set<int> st;
  vector<int> v1, v2;
  for (int k = n; k >= 1; k--) {
    if (v[k] == 0) {
      v2.push_back(k);
      for (int i = 0; i < v1.size(); i++) v2.push_back(v1[i]);
      v1.clear();
      v1 = v2;
      v2.clear();
    } else {
      int x = k + p;
      int f = 1;
      for (int i = 0; i < v1.size(); i++) {
        if (v[k] > 0 && v1[i] >= x) {
          v2.push_back(v1[i]);
          v[k]--;
        } else {
          v2.push_back(v1[i]);
        }
        if (v[k] == 0 && f) {
          v2.push_back(k);
          f = 0;
        }
      }
      v1.clear();
      v1 = v2;
      v2.clear();
    }
  }
  for (int i = 0; i < v1.size(); i++) cout << v1[i] << " ";
}
