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;
int B[1001];
int V[1001];
int A[1001];
bool MAP[1001];
int main() {
  int N, K;
  scanf("%d %d", &N, &K);
  for (int i = 0; i < N; i++) scanf("%d", &B[i]);
  for (int pos = 0; pos < N; pos++) {
    if (0) {
      printf("PRINT V\n");
      for (int j = 0; j < pos; j++) printf("%d ", V[j]);
      printf("\n");
    }
    for (int i = 1; i <= N; i++) {
      if (MAP[i]) continue;
      int cnt = 0;
      cnt = (int)(V + pos - lower_bound(V, V + pos, i + K));
      if (0) printf("COUNT for %d  with pos %d is %d\n", i, pos, cnt);
      if (cnt == B[i - 1]) {
        A[pos] = i;
        MAP[i] = 1;
        if (0) printf("FOR pos = %d found i = %d\n", pos + 1, i);
        int key = i, j;
        for (j = 0; j < pos; j++)
          if (V[j] > key) break;
        for (int k = pos; k > j; k--) V[k] = V[k - 1];
        V[j] = key;
        break;
      }
    }
  }
  for (int i = 0; i < N; i++) printf("%d ", A[i]);
  printf("\n");
  return 0;
}
