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 ti[1010], tree[1010];
int ans[1010];
int n, k;
void ins(int x) {
  if (x <= 0) return;
  for (; 0 < x; x -= x & -x) ++tree[x];
}
int find(int x) {
  int tot = 0;
  for (; x <= n; x += x & -x) tot += tree[x];
  return tot;
}
void solve() {
  memset(tree, 0, sizeof(tree));
  for (int i = 1; i <= n; ++i)
    for (int cur, j = 1; j <= n; ++j) {
      if (ti[j] == -1) continue;
      cur = find(j);
      if (cur == ti[j]) {
        ans[i] = j;
        ti[j] = -1;
        ins(j - k);
        break;
      }
    }
  printf("%d", ans[1]);
  for (int i = 2; i <= n; ++i) printf(" %d", ans[i]);
  puts("");
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &ti[i]);
  solve();
  return 0;
}
