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;
int b[1005];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> b[i];
  vector<int> v;
  for (int i = 1; i <= n; i++)
    if (b[i] == 0) v.emplace_back(i);
  for (int i = n; i >= 1; i--) {
    if (b[i] == 0) continue;
    int cnt = 0, q = 0;
    for (vector<int>::iterator x = v.begin(); x != v.end(); x++) {
      if (cnt == b[i]) {
        v.insert(x, i);
        q = 1;
        break;
      }
      if (i + k <= (*x)) cnt++;
    }
    if (q == 0 && cnt == b[i]) v.emplace_back(i);
  }
  for (auto x : v) printf("%d ", x);
}
