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: Time Limit Exceeded)

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <iostream>
#include <istream>
#include <iterator>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
//#include <tuple>
#include <iomanip>
#include <cstring>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<ll> vec;
typedef vector<vec> mat;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define revrep(i, n) for(ll i = (n-1); i >= 0; i--)
#define pb push_back
#define f first
#define s second
ll max(ll a, ll b){return (a > b) ? a : b;}
ll min(ll a, ll b){return (a < b) ? a : b;}
ll max3(ll a, ll b, ll c){return max(a, max(b, c));};
ll min3(ll a, ll b, ll c){return min(a, min(b, c));};
ll max4(ll a, ll b, ll c, ll d){return max(max(a, b), min(c, d));};
ll min4(ll a, ll b, ll c, ll d){return min(min(a, b), min(c, d));};
ll max5(ll a, ll b, ll c, ll d, ll e){return max(max(a, b), max3(c, d, e));};
ll min5(ll a, ll b, ll c, ll d, ll e){return min(min(a, b), min3(c, d, e));};

const ll INFL = 1LL << 60;//10^18 = 2^60
const int INF = 1 << 30;//10^9
ll MOD = 1000000007;
//ll MOD = 998244353;
/*
vector<ll> dy = {0, 0, 1, -1, 1, 1, -1, -1, 0};
vector<ll> dx = {1, -1, 0, 0, 1, -1, 1, -1, 0};
*/

ll pow_long(ll x, ll k){
  ll res = 1;
  while(k > 0){
    if(k % 2) res *= x;
    x *= x;
    k /= 2;
  }
  return res;
}
ll pow_mod(ll x, ll k){
  x %= MOD; x += MOD; x %= MOD;
  ll res = 1;
  while(k > 0){
    if(k % 2){
      res *= x; res %= MOD;
    }
    x *= x; x %= MOD;
    k /= 2;
  }
  return res;
}

ll inverse(ll x){return pow_mod(x, MOD - 2);};

ll gcd(ll a, ll b){
    if(b == 0) return a;
    return gcd(b, a % b);
}

ll lcm(ll x, ll y){return x / gcd(x, y) * y;};

ll kai_mod(ll x){
  if(x == 0) return 1;
  return x * kai_mod(x-1) % MOD;
}

/*
//コンビネーション
const int MAXcomb = 200010;
ll fac[MAXcomb], finv[MAXcomb], inv[MAXcomb];
//facはn!,finvは1/n!
//invは逆元
void COMinit(){
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for(int i = 2; i < MAXcomb; i++){
        fac[i] = fac[i-1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
        finv[i] = finv[i-1] * inv[i] % MOD;
    }
}
ll comb(int n, int k){
    if(n < k) return 0;
    if(n < 0 || k < 0) return 0;
    return fac[n] * finv[k] % MOD * finv[n-k] % MOD;
}
*/

ll N, K;
vector<ll> tree[100010];
vector<ll> depth(100010);
map<pair<ll, ll>, ll> memo;
ll ans;

void make_depth(ll now, ll dep){
  depth[now] = dep;
  for(ll to : tree[now]){
    make_depth(to, dep + 1);
  }
}

ll solve(ll now, ll dep){
  if(memo[{now, dep}] != 0) return (memo[{now, dep}] - 1);
  if(dep > K){
    ll res = 0;
    res++;
    for(ll to : tree[now]) res += solve(to, 2);
    memo[{now, dep}] = res + 1;
    return res;
  }else{
    ll res1 = 0;
    res1++;
    for(ll to : tree[now]) res1 += solve(to, 2);

    ll res2 = 0;
    for(ll to : tree[now]) res2 += solve(to, dep + 1);
    ll res = min(res1, res2);
    memo[{now, dep}] = res + 1;
    return res;
  }
}

int main(){
  cin >> N >> K;

  rep(i, N){
    ll a;
    cin >> a;
    a--;
    if(i == 0){
      if(a != 0) ans++;
      continue;
    }
    tree[a].pb(i);
  }
  make_depth(0, 0);
  ans += solve(0, 0);
  cout << ans << endl;

}