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 node {
  node() {}
  node(int _v) { v = _v; }
  int v;
};
int n, k;
int no[1000 + 100];
bool nav[1000 + 100];
bool operator<(const node& a, const node& b) {
  if (no[a.v] != no[b.v]) return no[a.v] < no[b.v];
  return a.v < b.v;
}
set<node> me;
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> no[i];
  for (int i = 0; i < n; i++) me.insert(node(i));
  for (int i = 0; i < n; i++) {
    int t = me.begin()->v;
    me.erase(me.begin());
    nav[t] = 1;
    cout << t + 1 << " ";
    for (int j = 0; j < n; j++)
      if (!nav[j] && no[j] && j + k <= t) {
        me.erase(node(j));
        no[j]--;
        me.insert(node(j));
      }
  }
  cout << endl;
  return 0;
}
