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 long long INF = 1e9 + 7;
const int N = 5e3 + 10;
int a[N];
int flag[N];
int sum[N];
int main() {
  int n, k;
  cin >> n >> k;
  priority_queue<int, vector<int>, greater<int>> pq;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 0) pq.push(i);
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++) {
    if (flag[i]) continue;
    while (a[i] != 0) {
      int x = pq.top();
      pq.pop();
      ans.push_back(x);
      flag[x] = 1;
      for (int j = i; j <= x; j++) {
        if (flag[j]) continue;
        if (j + k <= x) {
          a[j]--;
          if (a[j] == 0) pq.emplace(j);
        } else
          break;
      }
    }
    ans.push_back(pq.top());
    pq.pop();
    flag[i] = 1;
  }
  for (auto& x : ans) printf("%d ", x);
  return 0;
}
