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;
struct el {
  int cnt;
  int before;
  int val;
  bool operator<(el const& e) const {
    if (cnt != e.cnt)
      return cnt < e.cnt;
    else
      return before > e.before;
  }
};
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> b(n + 1);
  for (int i = 1; i <= n; i++) cin >> b[i];
  vector<el> v(n);
  for (int i = 1; i <= n; i++) {
    el e;
    e.cnt = b[i];
    e.before = i;
    e.val = i + k;
    v[i - 1] = e;
  }
  sort(v.begin(), v.end());
  vector<int> a(n + 1, -1);
  int num = 0;
  for (int i = 0; i < v.size(); i++) {
    int cnt = 0;
    for (int j = 1; j <= n; j++) {
      if (cnt == v[i].cnt) {
        for (int c = num + 1; c > j; c--) a[c] = a[c - 1];
        a[j] = v[i].before;
        num++;
        break;
      } else {
        if (a[j] >= v[i].val) cnt++;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (i > 1) cout << " ";
    cout << a[i];
  }
  cout << endl;
  return 0;
}
