#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k, l, r, z;
  cin >> n >> k;
  int *a = new int[n];
  int *b = new int[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = 0;
    if (a[i] == 0) b[i] = i - max(0, i - k) + 1 + min(n - 1, i + k) - i;
  }
  for (int i = 0; i < n; i++) {
    if (b[i] == 0)
      if (a[i] + k < i - k)
        b[i] = b[a[i] - 1] + i - max(0, i - k) + 1 + min(n - 1, i + k) - i;
      else {
        int l1 = max(0, a[i] - 1 - k), l2 = max(0, i - k);
        int r1 = min(n - 1, a[i] - 1 + k), r2 = min(n - 1, i + k);
        if (r1 < i) b[i] = b[a[i] - 1] + i - r1 + r2 - i;
        if (r1 == i) b[i] = b[a[i] - 1] + i - r1 + r2 - i;
        if (r1 > i) {
          b[i] = b[a[i] - 1] + r2 - r1;
        }
      }
  }
  for (int i = 0; i < n; i++) cout << b[i] << endl;
}

### Prompt

Create a solution in CPP for 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 main() {
  int n, k, l, r, z;
  cin >> n >> k;
  int *a = new int[n];
  int *b = new int[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = 0;
    if (a[i] == 0) b[i] = i - max(0, i - k) + 1 + min(n - 1, i + k) - i;
  }
  for (int i = 0; i < n; i++) {
    if (b[i] == 0)
      if (a[i] + k < i - k)
        b[i] = b[a[i] - 1] + i - max(0, i - k) + 1 + min(n - 1, i + k) - i;
      else {
        int l1 = max(0, a[i] - 1 - k), l2 = max(0, i - k);
        int r1 = min(n - 1, a[i] - 1 + k), r2 = min(n - 1, i + k);
        if (r1 < i) b[i] = b[a[i] - 1] + i - r1 + r2 - i;
        if (r1 == i) b[i] = b[a[i] - 1] + i - r1 + r2 - i;
        if (r1 > i) {
          b[i] = b[a[i] - 1] + r2 - r1;
        }
      }
  }
  for (int i = 0; i < n; i++) cout << b[i] << endl;
}

```