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 int mxn = 1024;
int n, k;
int a[mxn];
int main(int argc, char** argv) {
  scanf("%d", &n);
  scanf("%d", &k);
  priority_queue<int, vector<int>, greater<int> > q;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 0) {
      q.push(i);
    }
  }
  int printed = 0;
  while (q.size() > 0) {
    int z = q.top();
    q.pop();
    printf("%d", z);
    if (++printed == n) {
      puts("");
    } else {
      printf(" ");
    }
    int hi = z - k;
    for (int i = 1; i <= hi; i++) {
      if (--a[i] == 0) {
        q.push(i);
      }
    }
  }
  return 0;
}
