#include <bits/stdc++.h>
using namespace std;
const long double pi = 2 * acos(0.0);
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
  return a * (b / gcd(a, b));
}
template <class T, class TT>
bool pal(T a, TT n) {
  int k = 0;
  for (int i = 0; i <= n / 2; i++) {
    if (a[i] != a[n - i - 1]) {
      k = 1;
      break;
    }
  }
  return k ? 0 : 1;
}
int a[222222], dp[222222];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= k; i++) {
    int l = i - k;
    if (l < 1) l = 1;
    int r = i + k;
    if (r > n) r = n;
    dp[i] = r - l + 1;
  }
  for (int i = k + 1; i <= n; i++) {
    int t = a[i];
    if (i - a[i] - 1 < 2 * k && a[i]) {
      int l = a[i] + k;
      int r = i + k;
      if (l > n) l = n;
      if (r > n) r = n;
      dp[i] = dp[a[i]] + r - l;
    } else {
      int l = i - k;
      int r = i + k;
      if (l < 1) l = 1;
      if (r > n) r = n;
      dp[i] = dp[a[i]] + r - l + 1;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", dp[i]);
  getchar();
  getchar();
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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 namespace std;
const long double pi = 2 * acos(0.0);
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
  return a * (b / gcd(a, b));
}
template <class T, class TT>
bool pal(T a, TT n) {
  int k = 0;
  for (int i = 0; i <= n / 2; i++) {
    if (a[i] != a[n - i - 1]) {
      k = 1;
      break;
    }
  }
  return k ? 0 : 1;
}
int a[222222], dp[222222];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= k; i++) {
    int l = i - k;
    if (l < 1) l = 1;
    int r = i + k;
    if (r > n) r = n;
    dp[i] = r - l + 1;
  }
  for (int i = k + 1; i <= n; i++) {
    int t = a[i];
    if (i - a[i] - 1 < 2 * k && a[i]) {
      int l = a[i] + k;
      int r = i + k;
      if (l > n) l = n;
      if (r > n) r = n;
      dp[i] = dp[a[i]] + r - l;
    } else {
      int l = i - k;
      int r = i + k;
      if (l < 1) l = 1;
      if (r > n) r = n;
      dp[i] = dp[a[i]] + r - l + 1;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", dp[i]);
  getchar();
  getchar();
  return 0;
}

```