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.

Examples

Input

3 1
2 3 1


Output

2


Input

4 2
1 1 2 2


Output

0


Input

8 2
4 1 2 3 1 2 3 4


Output

3
from heapq import heappop,heappush

n,k = map(int,input().split())
a = [0] + list(map(int,input().split()))

ans = 0
if(a[1] != 1):
    ans += 1

links_c = [set() for _ in range(n+1)]
for i,ai in enumerate(a[2:],2):
    links_c[ai].add(i)

depth = [-1] * (n+1)
depth[1] = 0
stack = [1]
hq = []
while(stack):
    next = []
    while(stack):
        i = stack.pop()
        for j in links_c[i]:
            depth[j] = depth[i] + 1
            heappush(hq, (depth[j]*-1,j))
            next.append(j)
    stack = next[::]

while( hq[0][0]*-1 > k ):
    di,i = heappop(hq)
    if( depth[i] <= k ):
        continue

    for j in range(k-1):
        i = a[i]

    ans += 1
    links_c[a[i]].remove(i)
    stack = [i]
    while(stack):
        next = []
        while(stack):
            i = stack.pop()
            depth[i] = 1
            for j in links_c[i]:
                next.append(j)
        stack = next[::]

print(ans)


'''
首都の自己ループを除くと、首都を根として、
子→親の有向木になっている。

・首都のテレポート先は首都でないとダメ
→　長さ2以上の閉路を作るとうまくいかない

・首都以外は、首都までの距離をk以下にできればよい

もっとも根からの距離が遠い頂点xをとる。
xが根からの距離k以下　→　終了してOK
xが根からの距離k以上
→　xからk-1個離れた親を首都につなぎなおす
　そして再度最も遠い頂点を探す、を繰り返す。

'''