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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

//#pragma optimization_level 3

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <chrono>
#include <ctime>
#include <queue>
#include <math.h>
#include <deque>
#include <stack>
#include <iomanip>
#include <assert.h>
#include <stdio.h>
#include <cstring>
#include <random>

using namespace std;

#define int long long
#define ld long double
#define ull unsigned long long
//#define endl '\n'
#define try tr
#define pii pair<int, int>
#define ti tuple <int, int, int>
#define vi vector <int>
#define vpi vector <pii>
#define umap unordered_map
#define uset unordered_set
#define left left11
#define right right11
#define mp make_pair
#define all(c) (c).begin(), (c).end()
#define pb push_back
#define sqr(a) ((a) * (a))
#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

#define X first
#define Y second

ld TL = 1.0;
bool is_tl = 0;
int CALLS = 0;

inline bool IS() {
    if (++CALLS == 1000) {
        CALLS = 0;
        is_tl |= clock() > (TL - 0.1) * CLOCKS_PER_SEC;
    }
    return is_tl;
}

template<typename T1, typename T2>
inline void amin(T1 &a, T2 b) { if (a > b) a = b; }

template<typename T1, typename T2>
inline void amax(T1 &a, T2 b) { if (a < b) a = b; }

template<typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
    os << p.first << ' ' << p.second << endl;
    return os;
}

template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
    is >> p.first >> p.second;
    return is;
}

template<typename T>
ostream &operator<<(ostream &os, vector<T> &v) {
    for (auto &u : v) os << u << ' ';
    os << endl;
    return os;
}

mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const int N = 2e5 + 7, NS = 2e3 + 7, lg = 19, sq = 550, inf = 1e18 + 7, SM = 1e2 + 7;
const ld eps = 1e-9;
int hash_mod[4] = {1000000007, 998244353, 1000000009, 999999937}, mod = hash_mod[rnd() % 4];
int hash_pows[4] = {179, 239, 1007, 2003}, P = hash_pows[rnd() % 4];
vi v[N];
void solve() {
    int n, p, ans = 0, k;
    cin >> n >> k;
    for (int i = 0; i < n; i++) {
        cin >> p;
        p--;
        if(!i && p) {
            ans++;
            p = 0;
        }
        v[p].pb(i);
    }
    vi d(n, inf);
    d[0] = 0;
    set <pii> s;
    for (int i = 0; i < n; i++) {
        s.insert({d[i], i});
    }
    while(!s.empty()) {
        int i = s.begin() -> second;
        s.erase(s.begin());
       // cout << i + 1 << ' ' << d[i] << endl;
        if(d[i] > k) {
            d[i] = 0;
            ans++;
        }
        for (auto u : v[i]) {
            if(d[u] > d[i] + 1) {
                if(s.find({d[u], u}) != s.end())
                    s.erase({d[u], u});
                d[u] = d[i] + 1;
                s.insert({d[u], u});
            }
        }
    }
    cout << ans;
}

signed main() {
    fast
    solve();

}