#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;
const int MAXM = 2e6 + 10;
const long long MOD = 1e9 + 7;
const long long INF = 0x3f3f3f3f;
char rubbish[110];
long long read() {
  long long x = 0, f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch < '0' || ch > '9');
  do {
    x = x * 10 + ch - 48;
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return x * f;
}
int N, M, K;
int a[MAXN];
int dp[MAXN];
void init() {}
void solve() {
  int l1, r1, r2;
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; i++) {
    a[i] = read();
  }
  dp[0] = 0;
  for (int i = 1; i <= N; i++) {
    l1 = max(i - K, 1);
    r1 = min(i + K, N);
    dp[i] = dp[a[i]];
    if (a[i]) {
      r2 = min(a[i] + K, N) + 1;
    } else {
      r2 = 0;
    }
    dp[i] += r1 - max(l1, r2) + 1;
  }
  for (int i = 1; i <= N; i++) {
    printf("%d%c", dp[i], i == N ? '\n' : ' ');
  }
}
int main() {
  int T;
  T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 MAXN = 2e5 + 10;
const int MAXM = 2e6 + 10;
const long long MOD = 1e9 + 7;
const long long INF = 0x3f3f3f3f;
char rubbish[110];
long long read() {
  long long x = 0, f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch < '0' || ch > '9');
  do {
    x = x * 10 + ch - 48;
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return x * f;
}
int N, M, K;
int a[MAXN];
int dp[MAXN];
void init() {}
void solve() {
  int l1, r1, r2;
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; i++) {
    a[i] = read();
  }
  dp[0] = 0;
  for (int i = 1; i <= N; i++) {
    l1 = max(i - K, 1);
    r1 = min(i + K, N);
    dp[i] = dp[a[i]];
    if (a[i]) {
      r2 = min(a[i] + K, N) + 1;
    } else {
      r2 = 0;
    }
    dp[i] += r1 - max(l1, r2) + 1;
  }
  for (int i = 1; i <= N; i++) {
    printf("%d%c", dp[i], i == N ? '\n' : ' ');
  }
}
int main() {
  int T;
  T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```