#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
template <class T>
istream &operator>>(istream &is, vector<T> &v) {
  for (int i = 0; i < (int)v.size(); i++) is >> v[i];
  return is;
}
template <class T>
ostream &operator<<(ostream &os, vector<T> v) {
  for (int i = 0; i < (int)v.size(); i++) os << v[i] << ' ';
  os << '\n';
  return os;
}
char dcase(char a) {
  if (a >= 'A' && a <= 'Z') return a - 'A' + 'a';
  return a;
}
bool interchangable(char a, char b, const string &s) {
  bool af = false, bf = false;
  for (int i = 0; i < s.size(); i++) {
    af |= (a == s[i]);
    bf |= (b == s[i]);
  }
  return af && bf;
}
bool similar(const string &a, const string &b) {
  if (a.size() != b.size()) return false;
  bool sim = true;
  for (int i = 0; i < a.size() && sim; i++) {
    if (a[i] == b[i]) continue;
    if (interchangable(a[i], b[i], "0Oo") ||
        interchangable(a[i], b[i], "1lILi"))
      continue;
    if (dcase(a[i]) != dcase(b[i])) sim = false;
  }
  return sim;
}
signed main() {
  int n, k;
  cin >> n >> k;
  vector<int> p(n);
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  vector<int> ans(n, 0);
  for (int i = 0; i < n; i++) {
    if (p[i] == -1) {
      ans[i] = 1;
      ans[i] += min(k, i);
      ans[i] += min(k, n - i - 1);
      continue;
    }
    ans[i] = ans[p[i]] + min(k, i) + min(k, n - i - 1) + 1;
    if (min(n - 1, p[i] + k) >= max(0, i - k)) {
      ans[i] -= min(n - 1, p[i] + k) - max(0, (i - k)) + 1;
    }
  }
  cout << ans << '\n';
}

### Prompt

Please create a solution in CPP 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 GCC optimize("Ofast")
using namespace std;
template <class T>
istream &operator>>(istream &is, vector<T> &v) {
  for (int i = 0; i < (int)v.size(); i++) is >> v[i];
  return is;
}
template <class T>
ostream &operator<<(ostream &os, vector<T> v) {
  for (int i = 0; i < (int)v.size(); i++) os << v[i] << ' ';
  os << '\n';
  return os;
}
char dcase(char a) {
  if (a >= 'A' && a <= 'Z') return a - 'A' + 'a';
  return a;
}
bool interchangable(char a, char b, const string &s) {
  bool af = false, bf = false;
  for (int i = 0; i < s.size(); i++) {
    af |= (a == s[i]);
    bf |= (b == s[i]);
  }
  return af && bf;
}
bool similar(const string &a, const string &b) {
  if (a.size() != b.size()) return false;
  bool sim = true;
  for (int i = 0; i < a.size() && sim; i++) {
    if (a[i] == b[i]) continue;
    if (interchangable(a[i], b[i], "0Oo") ||
        interchangable(a[i], b[i], "1lILi"))
      continue;
    if (dcase(a[i]) != dcase(b[i])) sim = false;
  }
  return sim;
}
signed main() {
  int n, k;
  cin >> n >> k;
  vector<int> p(n);
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  vector<int> ans(n, 0);
  for (int i = 0; i < n; i++) {
    if (p[i] == -1) {
      ans[i] = 1;
      ans[i] += min(k, i);
      ans[i] += min(k, n - i - 1);
      continue;
    }
    ans[i] = ans[p[i]] + min(k, i) + min(k, n - i - 1) + 1;
    if (min(n - 1, p[i] + k) >= max(0, i - k)) {
      ans[i] -= min(n - 1, p[i] + k) - max(0, (i - k)) + 1;
    }
  }
  cout << ans << '\n';
}

```