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;
#pragma comment(linker, "/stack:16000000")
#pragma warning(disable : 4996)
const int inf = 1 << 25;
const double eps = 1e-9;
int main(int argc, char* argv[]) {
  int n, k;
  cin >> n >> k;
  vector<int> b(n);
  for (int i = (0); i < (n); ++i) cin >> b[i];
  vector<int> a(n);
  for (int i = (0); i < ((int)a.size()); ++i) a[i] = i + 1;
  int L = n + k;
  for (int i = n - 1; i >= 0; --i) {
    int j = i;
    while (b[i] > 0) {
      swap(a[j], a[j + 1]);
      if (a[j] >= L) --b[i];
      ++j;
    }
    --L;
  }
  for (int i = (0); i < ((int)a.size()); ++i) printf("%d ", a[i]);
  return 0;
}
