#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a;
  vector<long long> b;
  long long sum;
  if (k + 1 < n) {
    sum = k + 1;
    b.push_back(sum);
  } else {
    for (long long i = 0; i < n; i++) {
      cout << n << " ";
    }
    return 0;
  }
  for (long long i = 0; i < n; i++) {
    long long te;
    cin >> te;
    te--;
    a.push_back(te);
  }
  for (long long i = 1; i < n; i++) {
    long long re = 0;
    if (a[i] > -1) {
      if (a[i] + k > i) {
        if (a[i] + k < n) {
          re = a[i] + k - i;
        } else {
          re = n - i - 1;
        }
      }
      if (a[i] + k < i) {
        sum = 1;
      } else {
        sum = 0;
      }
      if (sum) {
        if (i - k > a[i] + k) {
          sum += k;
        } else {
          sum += i - 1 - (a[i] + k);
        }
      }
      sum += b[a[i]];
    } else {
      sum = 1;
      if (i - k > -1) {
        sum += k;
      } else {
        sum += i;
      }
    }
    if (i + k < n) {
      sum += k - re;
    } else {
      sum += n - 1 - i - re;
    }
    b.push_back(sum);
  }
  for (long long i = 0; i < n; i++) {
    cout << b[i] << " ";
  }
  return 0;
}

### Prompt

Your task is to create 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() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a;
  vector<long long> b;
  long long sum;
  if (k + 1 < n) {
    sum = k + 1;
    b.push_back(sum);
  } else {
    for (long long i = 0; i < n; i++) {
      cout << n << " ";
    }
    return 0;
  }
  for (long long i = 0; i < n; i++) {
    long long te;
    cin >> te;
    te--;
    a.push_back(te);
  }
  for (long long i = 1; i < n; i++) {
    long long re = 0;
    if (a[i] > -1) {
      if (a[i] + k > i) {
        if (a[i] + k < n) {
          re = a[i] + k - i;
        } else {
          re = n - i - 1;
        }
      }
      if (a[i] + k < i) {
        sum = 1;
      } else {
        sum = 0;
      }
      if (sum) {
        if (i - k > a[i] + k) {
          sum += k;
        } else {
          sum += i - 1 - (a[i] + k);
        }
      }
      sum += b[a[i]];
    } else {
      sum = 1;
      if (i - k > -1) {
        sum += k;
      } else {
        sum += i;
      }
    }
    if (i + k < n) {
      sum += k - re;
    } else {
      sum += n - 1 - i - re;
    }
    b.push_back(sum);
  }
  for (long long i = 0; i < n; i++) {
    cout << b[i] << " ";
  }
  return 0;
}

```