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 {
  long long int value;
  node *next;
};
node *root;
long long int k;
void add(long long int m, long long int n) {
  if (root == NULL) {
    root = new node();
    root->value = m;
    root->next = NULL;
  } else {
    node *temp, *crt, *i;
    temp = root;
    i = NULL;
    while (n != 0) {
      i = temp;
      if (temp->value >= m + k) {
        n--;
      }
      temp = temp->next;
    }
    if (temp == NULL) {
      crt = new node();
      crt->value = m;
      crt->next = NULL;
      i->next = crt;
    } else if (i == NULL) {
      crt = new node();
      crt->value = m;
      crt->next = root;
      root = crt;
    } else {
      crt = new node();
      crt->value = m;
      crt->next = temp;
      i->next = crt;
    }
  }
}
void print() {
  node *temp;
  temp = root;
  while (temp != NULL) {
    cout << temp->value << " ";
    temp = temp->next;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int n, i;
  cin >> n >> k;
  long long int ara[n];
  for (i = 0; i < n; i++) {
    cin >> ara[i];
  }
  root = NULL;
  for (i = n - 1; i >= 0; i--) {
    add(i + 1, ara[i]);
  }
  print();
}
