#include <bits/stdc++.h>
#pragma optimization_level 3
using namespace std;
vector<long long> v;
vector<long long> ans;
int main() {
  long long n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long aa;
    cin >> aa;
    v.push_back(aa);
  }
  for (long long i = 0; i < n; i++) {
    if (v[i] == 0) {
      ans.push_back(1 + min(k, i) + min(k, n - 1 - i));
      continue;
    }
    long long a = i + 1;
    long long cnt = ans[v[i] - 1];
    if (a - v[i] - 1 > 2 * k) {
      cnt += 1 + min(k, i) + min(k, n - 1 - i);
      ans.push_back(cnt);
      continue;
    }
    long long p = a - v[i] - 1 - min(k, n - 1 - v[i] + 1);
    p = 1 + min(k, n - 1 - i) + p;
    ans.push_back(cnt + p);
  }
  for (int ii = 0; ii < ans.size(); ii++) {
    cout << ans[ii] << " ";
  };
  return 0;
}

### Prompt

Generate 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>
#pragma optimization_level 3
using namespace std;
vector<long long> v;
vector<long long> ans;
int main() {
  long long n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long aa;
    cin >> aa;
    v.push_back(aa);
  }
  for (long long i = 0; i < n; i++) {
    if (v[i] == 0) {
      ans.push_back(1 + min(k, i) + min(k, n - 1 - i));
      continue;
    }
    long long a = i + 1;
    long long cnt = ans[v[i] - 1];
    if (a - v[i] - 1 > 2 * k) {
      cnt += 1 + min(k, i) + min(k, n - 1 - i);
      ans.push_back(cnt);
      continue;
    }
    long long p = a - v[i] - 1 - min(k, n - 1 - v[i] + 1);
    p = 1 + min(k, n - 1 - i) + p;
    ans.push_back(cnt + p);
  }
  for (int ii = 0; ii < ans.size(); ii++) {
    cout << ans[ii] << " ";
  };
  return 0;
}

```