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 a[1005];
int n, k, b[1005];
bool bio[1005];
int main(void) {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &b[i]);
  for (int i = 0; i < n; ++i) {
    bool flag = false;
    for (int x = 1; x <= k; ++x)
      if (!bio[x] && b[x] == 0) {
        printf("%d", x);
        bio[x] = true;
        flag = true;
        if (i + 1 < n)
          printf(" ");
        else
          printf("\n");
        break;
      }
    if (flag) continue;
    for (int x = 1; x + k <= n; ++x) {
      --b[x];
      if (bio[x + k]) continue;
      if (b[x + k] != 0) continue;
      printf("%d", x + k);
      bio[x + k] = true;
      if (i + 1 < n)
        printf(" ");
      else
        printf("\n");
      break;
    }
  }
  return 0;
}
