Problem:
Score : 800 points

Problem Statement

There are N towns in Snuke Kingdom, conveniently numbered 1 through N.
Town 1 is the capital.

Each town in the kingdom has a Teleporter, a facility that instantly transports a person to another place.
The destination of the Teleporter of town i is town a_i (1≤a_i≤N).
It is guaranteed that one can get to the capital from any town by using the Teleporters some number of times.

King Snuke loves the integer K.
The selfish king wants to change the destination of the Teleporters so that the following holds:

Starting from any town, one will be at the capital after using the Teleporters exactly K times in total.

Find the minimum number of the Teleporters whose destinations need to be changed in order to satisfy the king's desire.

Constraints

2≤N≤10^5

1≤a_i≤N

One can get to the capital from any town by using the Teleporters some number of times.

1≤K≤10^9

Input

The input is given from Standard Input in the following format:

N K
a_1 a_2 ... a_N

Output

Print the minimum number of the Teleporters whose destinations need to be changed in order to satisfy King Snuke's desire.

Sample Input 1

3 1
2 3 1

Sample Output 1

2

Change the destinations of the Teleporters to a = (1,1,1).

Sample Input 2

4 2
1 1 2 2

Sample Output 2

0

There is no need to change the destinations of the Teleporters, since the king's desire is already satisfied.

Sample Input 3

8 2
4 1 2 3 1 2 3 4

Sample Output 3

3

For example, change the destinations of the Teleporters to a = (1,1,2,1,1,2,2,4).

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/* 
容易发现这么几个性质：
(1)1一定在基环树的环上。否则在环上的点无法到达1
(2)当环的大小n >= 2时，给每个点按照顺序在mod n意义下标号，那么i + K != j + K(mod n)   <=>   i != j(mod n)。所以环的大小 = 1
那么最终一定呈现这样的结构：1往1连一个自环，剩下的点形成以1为根的深度<= K的树
把1的边连向1得到一棵树，剩下的点从上往下做，当一个点深度> K时接向根
*/

#include <cstdio>

using namespace std;
const int Max_N(100050);

int N, K, Ans, Father[Max_N], Head[Max_N], To[Max_N], Next[Max_N];

void dfs(int u, int k)
{
	if (k > K)
		k = 1, ++Ans;
	for (int i = Head[u];i;i = Next[i])
		dfs(To[i], k + 1);
}

int main()
{
	scanf("%d%d", &N, &K);
	for (int i = 1;i <= N;++i)
	{
		scanf("%d", Father + i);
		if (i == 1)
		{
			if (Father[i] != 1)
				++Ans;
		}
		else
			To[i] = i, Next[i] = Head[Father[i]], Head[Father[i]] = i;
	}
	dfs(1, 0);
	printf("%d", Ans);
	return 0;
}