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;
vector<int> l[1111];
int main() {
  int n, k;
  int s[1111];
  cin >> n >> k;
  set<pair<int, int> > ss;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    pair<int, int> tmp;
    tmp.first = s[i];
    tmp.second = i + 1;
    ss.insert(tmp);
  }
  for (int i = 0; i < n; i++) {
    for (int u = i + 1 + k; u <= n; u++) l[u].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) {
    pair<int, int> p = *ss.begin();
    ss.erase(ss.begin());
    cout << p.second << ' ';
    for (int u = 0; u < l[p.second].size(); u++) {
      pair<int, int> fnd;
      fnd.first = s[l[p.second][u] - 1];
      fnd.second = l[p.second][u];
      if (ss.find(fnd) != ss.end()) {
        ss.erase(fnd);
        fnd.first--;
        ss.insert(fnd);
        s[fnd.second - 1]--;
      }
    }
  }
  return 0;
}
