#include <bits/stdc++.h>
using namespace std;
class BChat {
 public:
  int N, K;
  vector<int> ans, R, used;
  vector<vector<int>> G;
  void input(std::istream& in, std::ostream& out) {
    ios_base::sync_with_stdio(false);
    in >> N >> K;
    ans = vector<int>(N);
    R = vector<int>(N);
    used = vector<int>(N, 0);
    G = vector<vector<int>>(N, vector<int>());
    for (int(i) = (0); (i) < (N); ++(i)) {
      int a;
      in >> a, a--;
      ans[i] = min(N - 1, i + K) - max(0, i - K) + 1;
      R[i] = min(N - 1, i + K);
      if (a > -1) G[i].push_back(a);
    }
  }
  void dfs(int s) {
    used[s] = 1;
    for (auto u : G[s]) {
      if (not used[u]) dfs(u);
      if (s - K <= R[u])
        ans[s] = ans[u] + R[s] - R[u];
      else
        ans[s] += ans[u];
    }
  }
  void solve(std::istream& in, std::ostream& out) {
    input(in, out);
    for (int(i) = (0); (i) < (N); ++(i)) dfs(i);
    for (int(i) = (0); (i) < (N); ++(i)) out << ans[i] << ' ';
    out << '\n';
  }
};
int main() {
  BChat solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  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;
class BChat {
 public:
  int N, K;
  vector<int> ans, R, used;
  vector<vector<int>> G;
  void input(std::istream& in, std::ostream& out) {
    ios_base::sync_with_stdio(false);
    in >> N >> K;
    ans = vector<int>(N);
    R = vector<int>(N);
    used = vector<int>(N, 0);
    G = vector<vector<int>>(N, vector<int>());
    for (int(i) = (0); (i) < (N); ++(i)) {
      int a;
      in >> a, a--;
      ans[i] = min(N - 1, i + K) - max(0, i - K) + 1;
      R[i] = min(N - 1, i + K);
      if (a > -1) G[i].push_back(a);
    }
  }
  void dfs(int s) {
    used[s] = 1;
    for (auto u : G[s]) {
      if (not used[u]) dfs(u);
      if (s - K <= R[u])
        ans[s] = ans[u] + R[s] - R[u];
      else
        ans[s] += ans[u];
    }
  }
  void solve(std::istream& in, std::ostream& out) {
    input(in, out);
    for (int(i) = (0); (i) < (N); ++(i)) dfs(i);
    for (int(i) = (0); (i) < (N); ++(i)) out << ans[i] << ' ';
    out << '\n';
  }
};
int main() {
  BChat solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}

```