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 data[1000];
int n, p;
int iTree[1 << 11];
const int base = (1 << 10);
bool isUsed[1000];
int ans[1000];
int getVal(int ts, int te, int cs, int ce, int ind) {
  if (ts >= te || cs >= ce) return 0;
  if (ts >= ce || cs >= te) return 0;
  if (ts == cs && te == ce) return iTree[ind];
  int mid = (cs + ce) / 2;
  int ret = getVal(ts, min(te, mid), cs, mid, ind * 2);
  ret += getVal(max(ts, mid), te, mid, ce, ind * 2 + 1);
  return ret;
}
void update(int ind) {
  ind += base;
  iTree[ind] = 1;
  ind /= 2;
  while (ind) {
    iTree[ind] = iTree[ind * 2] + iTree[ind * 2 + 1];
    ind /= 2;
  }
}
int main(void) {
  scanf("%d %d", &n, &p);
  for (int i = 0; i < n; i++) scanf("%d", data + i);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (isUsed[j]) continue;
      int sum = getVal(j + p, n, 0, base, 1);
      if (sum == data[j]) {
        isUsed[j] = true;
        update(j);
        ans[i] = j + 1;
        break;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (i) printf(" ");
    printf("%d", ans[i]);
  }
  printf("\n");
  return 0;
}
