#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int MAX = 2e5 + 5;
const long long MOD = 1000000007;
const long long MOD2 = 2010405347;
const long long INF = 2e18;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
const double pi = acos(-1);
const double EPS = 1e-9;
const int block = 315;
inline long long pw(long long x, long long y, long long md = MOD) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % md;
    x = x * x % md, y >>= 1;
  }
  return ret;
}
inline void add(int &a, int b, int md = MOD) {
  a = a + b >= md ? a + b - md : a + b;
}
int n, k, x[MAX], dp[MAX];
int f(int nw) {
  if (dp[nw] != -1) return dp[nw];
  if (!x[nw]) return dp[nw] = 1 + min(k, nw - 1) + min(k, n - nw);
  int a = min(n, x[nw] + k), b = max(1, nw - k), c = min(n, nw + k);
  if (a < b) return dp[nw] = (c - b + 1) + f(x[nw]);
  return dp[nw] = (c - a) + f(x[nw]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = (1); i <= (n); ++i) cin >> x[i];
  memset(dp, -1, sizeof dp);
  for (int i = (1); i <= (n); ++i) cout << f(i) << ' ';
  cout << '\n';
  return 0;
}

### 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>
#pragma GCC optimize("O3")
#pragma GCC target("tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int MAX = 2e5 + 5;
const long long MOD = 1000000007;
const long long MOD2 = 2010405347;
const long long INF = 2e18;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
const double pi = acos(-1);
const double EPS = 1e-9;
const int block = 315;
inline long long pw(long long x, long long y, long long md = MOD) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % md;
    x = x * x % md, y >>= 1;
  }
  return ret;
}
inline void add(int &a, int b, int md = MOD) {
  a = a + b >= md ? a + b - md : a + b;
}
int n, k, x[MAX], dp[MAX];
int f(int nw) {
  if (dp[nw] != -1) return dp[nw];
  if (!x[nw]) return dp[nw] = 1 + min(k, nw - 1) + min(k, n - nw);
  int a = min(n, x[nw] + k), b = max(1, nw - k), c = min(n, nw + k);
  if (a < b) return dp[nw] = (c - b + 1) + f(x[nw]);
  return dp[nw] = (c - a) + f(x[nw]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = (1); i <= (n); ++i) cin >> x[i];
  memset(dp, -1, sizeof dp);
  for (int i = (1); i <= (n); ++i) cout << f(i) << ' ';
  cout << '\n';
  return 0;
}

```