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 <class T>
void show(T a, int n) {
  for (int i = 0; i < n; ++i) cout << a[i] << ' ';
  cout << endl;
}
template <class T>
void show(T a, int r, int l) {
  for (int i = 0; i < r; ++i) show(a[i], l);
  cout << endl;
}
const int N = 1 << 10;
const int M = 5000;
const int oo = 10000 * 10000 * 10;
int n, m;
int ni[N];
int a[N];
int main() {
  int i, j, cas = 0;
  scanf("%d %d", &n, &m);
  for (i = 1; i <= n; ++i) scanf("%d", &ni[i]);
  for (i = n; i >= 1; --i) {
    a[n + 1 - i] = i;
    j = n + 1 - i;
    int t = (0 > n - i - m + 1 ? 0 : n - i - m + 1);
    ni[i] = t - ni[i];
    while (j > 1) {
      if (a[j - 1] >= a[j] + m) {
        if (ni[i] > 0) {
          --ni[i];
          swap(a[j], a[j - 1]);
        } else
          break;
      } else
        swap(a[j], a[j - 1]);
      --j;
    }
  }
  show(a + 1, n);
  return 0;
}
