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 NMAX = 1010;
int n, k;
int a[NMAX], nxt[NMAX], b[NMAX];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> b[i];
  }
  int sz = 1;
  int cur = 0;
  for (int i = n; i > 0; --i) {
    int seen = 0;
    int prev = cur;
    while (seen < b[i]) {
      prev = nxt[prev];
      seen += a[prev] >= k + i;
    }
    nxt[sz] = nxt[prev];
    nxt[prev] = sz;
    a[sz] = i;
    ++sz;
  }
  cur = nxt[cur];
  while (cur != 0) {
    cout << a[cur] << " ";
    cur = nxt[cur];
  }
  cout << endl;
  return 0;
}
