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>
#pragma comment(linker, "/STACK:64000000")
using namespace std;
void __never(int a) { printf("\nOPS %d", a); }
int n, k;
int B[1024];
int A[1024];
void sol() {
  int a_sz = 1;
  A[1] = n;
  for (int a = (n - 1); a >= (1); a--) {
    int tmp = 0;
    for (int b = (0); b <= (a_sz); b++) {
      if (b)
        if (A[b] >= a + k) tmp++;
      if (tmp == B[a]) {
        for (int c = (a_sz); c >= (b + 1); c--) A[c + 1] = A[c];
        a_sz++;
        A[b + 1] = a;
        break;
      }
    }
  }
  for (int a = (1); a <= (n); a++) printf("%d ", A[a]);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int a = (1); a <= (n); a++) scanf("%d", &B[a]);
  sol();
  return 0;
}
