#include <bits/stdc++.h>
using namespace std;
int n, k;
int a[100000], vis[100000];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    a[i]--;
  }
  for (int i = 0; i < (n); i++) {
    int loReach = max(0, i - k);
    int hiReach = min(n - 1, i + k);
    if (a[i] == -1) {
      vis[i] = hiReach - loReach + 1;
    } else {
      int e = min(n - 1, a[i] + k);
      vis[i] = vis[a[i]] + hiReach - max(e, loReach - 1);
    }
  }
  for (int i = 0; i < (n); i++) cout << vis[i] << " ";
  return 0;
}

### 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 n, k;
int a[100000], vis[100000];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    a[i]--;
  }
  for (int i = 0; i < (n); i++) {
    int loReach = max(0, i - k);
    int hiReach = min(n - 1, i + k);
    if (a[i] == -1) {
      vis[i] = hiReach - loReach + 1;
    } else {
      int e = min(n - 1, a[i] + k);
      vis[i] = vis[a[i]] + hiReach - max(e, loReach - 1);
    }
  }
  for (int i = 0; i < (n); i++) cout << vis[i] << " ";
  return 0;
}

```