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>
int n, k;
int v[1005], sol[1005], f[1005];
int main() {
  scanf("%d %d", &n, &k);
  int i, j, ind, nr, l;
  for (i = 1; i <= n; i++) scanf("%d", &v[i]);
  for (i = n; i >= 1; i--) {
    if (v[i]) break;
    for (j = sol[0]; j >= 1; j--) sol[j + 1] = sol[j];
    sol[1] = i;
    sol[0]++;
  }
  for (; i >= 1; i--) {
    nr = 0;
    for (j = 0; j <= sol[0]; j++) {
      if (j && sol[j] >= i + k) nr++;
      if (nr == v[i]) {
        for (l = sol[0]; l > j; l--) sol[l + 1] = sol[l];
        sol[j + 1] = i;
        sol[0]++;
        break;
      }
    }
  }
  for (i = 1; i <= n; i++) printf("%d ", sol[i]);
  return 0;
}
