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 maxn = 1010;
int n, k;
int nxt[maxn];
pair<int, int> a[maxn];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i].first), a[i].second = i;
  sort(a + 1, a + 1 + n);
  int first = 0;
  for (int i = 1; i <= n; ++i) {
    int x = a[i].second, need = a[i].first;
    if (!first)
      first = x;
    else if (!need) {
      int j = first;
      while (nxt[j]) j = nxt[j];
      nxt[j] = x;
    } else {
      int j = first;
      while (1) {
        if (j >= x + k) --need;
        if (need == 0) {
          int mn = x;
          while (nxt[j])
            if (nxt[j] < x)
              j = nxt[j];
            else
              break;
          nxt[x] = nxt[j];
          nxt[j] = x;
          break;
        }
        j = nxt[j];
      }
    }
  }
  while (first) {
    printf("%d ", first);
    first = nxt[first];
  }
  return 0;
}
