#include <bits/stdc++.h>
using std::abs;
using std::array;
using std::cerr;
using std::cin;
using std::cout;
using std::generate;
using std::make_pair;
using std::map;
using std::max;
using std::max_element;
using std::min;
using std::min_element;
using std::pair;
using std::reverse;
using std::set;
using std::sort;
using std::string;
using std::unique;
using std::vector;
template <typename T>
T input() {
  T res;
  cin >> res;
  {};
  return res;
}
template <typename IT>
void input_seq(IT b, IT e) {
  std::generate(b, e,
                input<typename std::remove_reference<decltype(*b)>::type>);
}
int main() {
  std::iostream::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  input_seq(a.begin(), a.end());
  vector<int> dp(n);
  for (int i = 0; i != n; ++i) {
    if (a[i] == 0)
      dp[i] = min(i + k, n - 1) - max(0, i - k) + 1;
    else {
      dp[i] = dp[a[i] - 1];
      int L1 = max(0, a[i] - 1 - k);
      int R1 = min(n - 1, a[i] - 1 + k);
      int L2 = max(0, i - k);
      int R2 = min(n - 1, i + k);
      dp[i] += R2 - max(L2 - 1, R1);
    }
  }
  for (auto elem : dp) cout << elem << " ";
  cout << "\n";
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 std::abs;
using std::array;
using std::cerr;
using std::cin;
using std::cout;
using std::generate;
using std::make_pair;
using std::map;
using std::max;
using std::max_element;
using std::min;
using std::min_element;
using std::pair;
using std::reverse;
using std::set;
using std::sort;
using std::string;
using std::unique;
using std::vector;
template <typename T>
T input() {
  T res;
  cin >> res;
  {};
  return res;
}
template <typename IT>
void input_seq(IT b, IT e) {
  std::generate(b, e,
                input<typename std::remove_reference<decltype(*b)>::type>);
}
int main() {
  std::iostream::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  input_seq(a.begin(), a.end());
  vector<int> dp(n);
  for (int i = 0; i != n; ++i) {
    if (a[i] == 0)
      dp[i] = min(i + k, n - 1) - max(0, i - k) + 1;
    else {
      dp[i] = dp[a[i] - 1];
      int L1 = max(0, a[i] - 1 - k);
      int R1 = min(n - 1, a[i] - 1 + k);
      int L2 = max(0, i - k);
      int R2 = min(n - 1, i + k);
      dp[i] += R2 - max(L2 - 1, R1);
    }
  }
  for (auto elem : dp) cout << elem << " ";
  cout << "\n";
  return 0;
}

```