#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  int k;
  cin >> n >> k;
  int a;
  vector<pair<int, int> > v;
  int temp;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    temp = 1;
    temp += min(n - i, k);
    temp += min(k, i - 1);
    v.push_back(make_pair(a, temp));
  }
  for (int i = 0; i < n; i++) {
    if (v[i].first == 0) {
      cout << v[i].second << " ";
      continue;
    }
    temp = v[i].second + v[v[i].first - 1].second;
    if (i + 1 - k <= 1) {
      temp -= i;
    } else if (i + 1 - v[i].first <= 2 * k && i + 1 - k > v[i].first) {
      temp -= v[i].first + k - (i - k);
    } else if (i + 1 - k <= v[i].first) {
      temp -= k;
    }
    if (v[i].first + k >= n) {
      temp -= n - i;
    } else if (v[i].first + k >= i + 1) {
      temp -= v[i].first + k - i;
    }
    cout << temp << " ";
    v[i].second = temp;
  }
  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>
using namespace std;
int main() {
  int n;
  int k;
  cin >> n >> k;
  int a;
  vector<pair<int, int> > v;
  int temp;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    temp = 1;
    temp += min(n - i, k);
    temp += min(k, i - 1);
    v.push_back(make_pair(a, temp));
  }
  for (int i = 0; i < n; i++) {
    if (v[i].first == 0) {
      cout << v[i].second << " ";
      continue;
    }
    temp = v[i].second + v[v[i].first - 1].second;
    if (i + 1 - k <= 1) {
      temp -= i;
    } else if (i + 1 - v[i].first <= 2 * k && i + 1 - k > v[i].first) {
      temp -= v[i].first + k - (i - k);
    } else if (i + 1 - k <= v[i].first) {
      temp -= k;
    }
    if (v[i].first + k >= n) {
      temp -= n - i;
    } else if (v[i].first + k >= i + 1) {
      temp -= v[i].first + k - i;
    }
    cout << temp << " ";
    v[i].second = temp;
  }
  return 0;
}

```