#include <bits/stdc++.h>
using namespace std;
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:336777216")
inline int nxt() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long lnxt() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
const int base = (int)1e9;
const int N = (int)1e6 + 100;
const int MAXN = (int)1e3 + 100;
const int MOD = (int)1e9 + 7;
const int INF = INT_MAX;
const long long LL_INF = LLONG_MAX;
const long double PI = 3.1415926535897932384626433832795028841971;
const long double EPS = 1e-8;
long long cnt = 0, ans = 0;
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> V(n);
  for (int p = 0; p < n; ++p) {
    int to;
    cin >> to;
    int mn1 = min(k, p);
    int mn2 = min(k, n - p - 1);
    V[p] = mn1 + mn2;
    ++V[p];
    if (to) {
      int l = max(p - k + 1, 1), r = min(to + k, n);
      if (l <= r) {
        V[p] -= r - l + 1;
      }
    }
    if (to) {
      V[p] += V[to - 1];
    }
  }
  for (auto g : V) cout << g << ' ';
  cout << endl;
}
int main() {
  int T = 1;
  for (; T; --T) solve();
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:336777216")
inline int nxt() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long lnxt() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
const int base = (int)1e9;
const int N = (int)1e6 + 100;
const int MAXN = (int)1e3 + 100;
const int MOD = (int)1e9 + 7;
const int INF = INT_MAX;
const long long LL_INF = LLONG_MAX;
const long double PI = 3.1415926535897932384626433832795028841971;
const long double EPS = 1e-8;
long long cnt = 0, ans = 0;
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> V(n);
  for (int p = 0; p < n; ++p) {
    int to;
    cin >> to;
    int mn1 = min(k, p);
    int mn2 = min(k, n - p - 1);
    V[p] = mn1 + mn2;
    ++V[p];
    if (to) {
      int l = max(p - k + 1, 1), r = min(to + k, n);
      if (l <= r) {
        V[p] -= r - l + 1;
      }
    }
    if (to) {
      V[p] += V[to - 1];
    }
  }
  for (auto g : V) cout << g << ' ';
  cout << endl;
}
int main() {
  int T = 1;
  for (; T; --T) solve();
  return 0;
}

```