#include <bits/stdc++.h>
using namespace std;
const int N = 200002, M = 350;
mt19937 gen(time(NULL));
template <typename T = int>
inline T read() {
  T val = 0, sign = 1;
  char ch;
  for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') sign = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) val = val * 10 + ch - '0';
  return sign * val;
}
int a[N];
int ans[N];
void solve() {
  int n = read(), k = read();
  for (int i = 0; i < n; i++) a[i] = read() - 1;
  for (int i = 0; i < n; i++) {
    if (a[i] == -1) {
      ans[i] = 1 + min(i, k) + min(n - i - 1, k);
      continue;
    }
    ans[i] = 1 + min(n - i - 1, k);
    int delta_left = i - a[i] - 1;
    if (delta_left >= 2 * k)
      ans[i] += ans[a[i]] + k;
    else if (delta_left >= k)
      ans[i] += ans[a[i]] + delta_left - k;
    else {
      ans[i] -= (1 + min(n - i - 1, k));
      ans[i] += ans[a[i]];
      int lpoint = a[i] + k;
      int rpoint = i + min(n - i - 1, k);
      ans[i] += max(0, rpoint - lpoint);
    }
  }
  for (int i = 0; i < n; i++) printf("%d ", ans[i]);
}
void precalc() {}
signed main() {
  int t = 1;
  precalc();
  while (t--) {
    clock_t z = clock();
    solve();
    fprintf(stderr, "Total Time: %.3f\n",
            (double)(clock() - z) / CLOCKS_PER_SEC),
        fflush(stderr);
  }
}

### Prompt

Your challenge is to write 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;
const int N = 200002, M = 350;
mt19937 gen(time(NULL));
template <typename T = int>
inline T read() {
  T val = 0, sign = 1;
  char ch;
  for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') sign = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) val = val * 10 + ch - '0';
  return sign * val;
}
int a[N];
int ans[N];
void solve() {
  int n = read(), k = read();
  for (int i = 0; i < n; i++) a[i] = read() - 1;
  for (int i = 0; i < n; i++) {
    if (a[i] == -1) {
      ans[i] = 1 + min(i, k) + min(n - i - 1, k);
      continue;
    }
    ans[i] = 1 + min(n - i - 1, k);
    int delta_left = i - a[i] - 1;
    if (delta_left >= 2 * k)
      ans[i] += ans[a[i]] + k;
    else if (delta_left >= k)
      ans[i] += ans[a[i]] + delta_left - k;
    else {
      ans[i] -= (1 + min(n - i - 1, k));
      ans[i] += ans[a[i]];
      int lpoint = a[i] + k;
      int rpoint = i + min(n - i - 1, k);
      ans[i] += max(0, rpoint - lpoint);
    }
  }
  for (int i = 0; i < n; i++) printf("%d ", ans[i]);
}
void precalc() {}
signed main() {
  int t = 1;
  precalc();
  while (t--) {
    clock_t z = clock();
    solve();
    fprintf(stderr, "Total Time: %.3f\n",
            (double)(clock() - z) / CLOCKS_PER_SEC),
        fflush(stderr);
  }
}

```