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 int N = 1010;
const int BS = 5;
int n, k, a[N];
bool f[N];
int b[N];
int u[N];
void modify(int x, int v) {
  for (x += BS; x < N; x += x & -x) u[x] += v;
}
int getsum(int x) {
  int s = 0;
  for (x += BS; x; x -= x & -x) s += u[x];
  return s;
}
int check(int i) {
  if (i + k > n) return b[i] == 0;
  return getsum(n) - getsum(i + k - 1) == b[i];
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", b + i);
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (f[j]) continue;
      if (check(j)) {
        a[i] = j;
        f[j] = 1;
        modify(j, 1);
        break;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    printf("%d%c", a[i], " \n"[i == n - 1]);
  }
  return 0;
}
