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[1002] = {0}, A[1002] = {0}, Min[1002] = {0};
bool used[1002] = {0};
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &B[i]);
  Min[0] = 1;
  B[0] = 1;
  for (int i = 1; i <= n; i++) Min[i] = min(Min[i - 1], B[i]);
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= n; j++) {
      if (used[j]) continue;
      if (B[j] == 0 && (j - k <= 0 || Min[j - k] > 0)) {
        A[i] = j;
        used[j] = true;
        for (int t = 0; t + k + 1 <= j; t++) {
          if (!used[t + 1]) {
            B[t + 1]--;
          } else
            Min[t + 1] = Min[t];
        }
        break;
      }
    }
    for (int t = 0; t < n; t++) {
      if (!used[t + 1]) {
        if (t == 0)
          Min[t + 1] = B[t + 1];
        else
          Min[t + 1] = min(Min[t], B[t + 1]);
      } else
        Min[t + 1] = Min[t];
    }
  }
  for (int i = 0; i < n; i++) {
    if (i != 0) cout << " ";
    cout << A[i];
  }
  cout << endl;
  return 0;
}
