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;
template <typename TYPE>
inline TYPE sqr(const TYPE& a) {
  return (a) * (a);
}
const int INF = 1000 * 1000 * 1000;
const long double EPS = 1e-9;
const long double PI = 2 * acos(0.0);
const int N = 1010;
int n, k;
int b[N];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < int(n); ++i) scanf("%d", &b[i]);
  vector<int> a;
  while (a.size() < n) {
    for (int i = n - 1; i >= 0; --i) {
      int cnt = b[i];
      if (cnt == 0) {
        a.insert(a.begin(), i);
        continue;
      }
      {
        for (int j = 0; j < int(a.size()); ++j) {
          if (a[j] >= i + k) {
            --cnt;
            if (cnt <= 0) {
              int t = j + 1;
              a.insert(a.begin() + t, i);
              b[i] = -1;
              break;
            }
          }
        }
      }
    }
  }
  for (int i = 0; i < int(a.size()); ++i) printf("%d ", a[i] + 1);
  printf("\n");
  return 0;
}
