#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
bool umin(T1& x, const T2& y) {
  return (x > y ? (x = y, true) : false);
}
template <typename T1, typename T2>
bool umax(T1& x, const T2& y) {
  return (x < y ? (x = y, true) : false);
}
template <typename T>
void read(T& x) {
  char ch;
  T p = 1;
  x = 0;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') p = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= p;
}
void read(string& x) {
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  while (ch > ' ') x += ch, ch = getchar();
}
void read(char& x) {
  do {
    x = getchar();
  } while (x <= ' ');
}
template <typename T, typename... R>
void read(T& x, R&... y) {
  read(x);
  read(y...);
}
const int N = (int)1e6 + 3;
const int INF = (int)1e9 + 17;
const long long mod = (long long)1e9 + 7;
const long long LLINF = (long long)1e18 + 17;
int n, m;
int p[N];
int ans[N];
int main() {
  read(n, m);
  for (int i = (1); i < (n + 1); i++) read(p[i]);
  for (int i = (1); i < (n + 1); i++) {
    ans[i] = ans[p[i]];
    if (p[i] == 0) p[i] -= m;
    int tl = max(i - m - 1, min(n, p[i] + m));
    ans[i] += min(i + m, n) - tl;
    printf("%d%c", ans[i], " \n"[i == n]);
  }
  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;
template <typename T1, typename T2>
bool umin(T1& x, const T2& y) {
  return (x > y ? (x = y, true) : false);
}
template <typename T1, typename T2>
bool umax(T1& x, const T2& y) {
  return (x < y ? (x = y, true) : false);
}
template <typename T>
void read(T& x) {
  char ch;
  T p = 1;
  x = 0;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') p = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= p;
}
void read(string& x) {
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  while (ch > ' ') x += ch, ch = getchar();
}
void read(char& x) {
  do {
    x = getchar();
  } while (x <= ' ');
}
template <typename T, typename... R>
void read(T& x, R&... y) {
  read(x);
  read(y...);
}
const int N = (int)1e6 + 3;
const int INF = (int)1e9 + 17;
const long long mod = (long long)1e9 + 7;
const long long LLINF = (long long)1e18 + 17;
int n, m;
int p[N];
int ans[N];
int main() {
  read(n, m);
  for (int i = (1); i < (n + 1); i++) read(p[i]);
  for (int i = (1); i < (n + 1); i++) {
    ans[i] = ans[p[i]];
    if (p[i] == 0) p[i] -= m;
    int tl = max(i - m - 1, min(n, p[i] + m));
    ans[i] += min(i + m, n) - tl;
    printf("%d%c", ans[i], " \n"[i == n]);
  }
  return 0;
}

```