#include <bits/stdc++.h>
using namespace std;
int sum(int *a, int k, int n, int start) {
  int res = 0;
  int i = 0;
  int local = 0;
  while (start != 0) {
    if (start - k > 0) {
      if (n < start - k)
        ;
      else if (n > start + k) {
        res += 2 * k + 1;
      } else if (start - k <= n && start + k >= n) {
        res += n - (start - k) + 1;
      }
    } else {
      if (n > start + k) {
        res += start + k;
      } else {
        res += n > 0 ? n : 0;
      }
    }
    n = start - k - 1;
    if (n <= 0) break;
    start = a[start];
  }
  return res;
}
int main() {
  int n = 0;
  int k = 0;
  scanf("%d", &n);
  scanf("%d", &k);
  int *a = new int[n + 1];
  for (int i = 1; i < n + 1; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i < n + 1; i++) {
    printf("%d ", sum(a, k, n, i));
  }
}

### Prompt

Please formulate a Cpp solution to the following problem:

There are times you recall a good old friend and everything you've come through together. Luckily there are social networks — they store all your message history making it easy to know what you argued over 10 years ago.

More formal, your message history is a sequence of messages ordered by time sent numbered from 1 to n where n is the total number of messages in the chat.

Each message might contain a link to an earlier message which it is a reply to. When opening a message x or getting a link to it, the dialogue is shown in such a way that k previous messages, message x and k next messages are visible (with respect to message x). In case there are less than k messages somewhere, they are yet all shown.

Digging deep into your message history, you always read all visible messages and then go by the link in the current message x (if there is one) and continue reading in the same manner.

Determine the number of messages you'll read if your start from message number t for all t from 1 to n. Calculate these numbers independently. If you start with message x, the initial configuration is x itself, k previous and k next messages. Messages read multiple times are considered as one.

Input

The first line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ n) — the total amount of messages and the number of previous and next messages visible.

The second line features a sequence of integers a1, a2, ..., an (0 ≤ ai < i), where ai denotes the i-th message link destination or zero, if there's no link from i. All messages are listed in chronological order. It's guaranteed that the link from message x goes to message with number strictly less than x.

Output

Print n integers with i-th denoting the number of distinct messages you can read starting from message i and traversing the links while possible.

Examples

Input

6 0
0 1 1 2 3 2


Output

1 2 2 3 3 3 


Input

10 1
0 1 0 3 4 5 2 3 7 0


Output

2 3 3 4 5 6 6 6 8 2 


Input

2 2
0 1


Output

2 2 

Note

Consider i = 6 in sample case one. You will read message 6, then 2, then 1 and then there will be no link to go.

In the second sample case i = 6 gives you messages 5, 6, 7 since k = 1, then 4, 5, 6, then 2, 3, 4 and then the link sequence breaks. The number of distinct messages here is equal to 6.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int sum(int *a, int k, int n, int start) {
  int res = 0;
  int i = 0;
  int local = 0;
  while (start != 0) {
    if (start - k > 0) {
      if (n < start - k)
        ;
      else if (n > start + k) {
        res += 2 * k + 1;
      } else if (start - k <= n && start + k >= n) {
        res += n - (start - k) + 1;
      }
    } else {
      if (n > start + k) {
        res += start + k;
      } else {
        res += n > 0 ? n : 0;
      }
    }
    n = start - k - 1;
    if (n <= 0) break;
    start = a[start];
  }
  return res;
}
int main() {
  int n = 0;
  int k = 0;
  scanf("%d", &n);
  scanf("%d", &k);
  int *a = new int[n + 1];
  for (int i = 1; i < n + 1; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i < n + 1; i++) {
    printf("%d ", sum(a, k, n, i));
  }
}

```