Problem:
Score: 400 points

Problem Statement

The Kingdom of Takahashi has N towns, numbered 1 through N.

There is one teleporter in each town. The teleporter in Town i (1 \leq i \leq N) sends you to Town A_i.

Takahashi, the king, loves the positive integer K. The selfish king wonders what town he will be in if he starts at Town 1 and uses a teleporter exactly K times from there.

Help the king by writing a program that answers this question.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq A_i \leq N

1 \leq K \leq 10^{18}

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \dots A_N

Output

Print the integer representing the town the king will be in if he starts at Town 1 and uses a teleporter exactly K times from there.

Sample Input 1

4 5
3 2 4 1

Sample Output 1

4

If we start at Town 1 and use the teleporter 5 times, our travel will be as follows: 1 \to 3 \to 4 \to 1 \to 3 \to 4.

Sample Input 2

6 727202214173249351
6 5 2 5 3 2

Sample Output 2

2

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
constexpr int Inf = 1000000000;
constexpr ll INF= 1e18;
constexpr ll MOD = 1000000007;
const double PI = 3.1415926535897;
typedef pair<ll,ll> P;

template<typename T>
T Pow(T a,T b) {
    T ret = 1;
    for(int i = 0;i < b;i++) {
        ret *= a;
    }
    return ret;
}

ll mod(ll val) {
    ll res = val % MOD;
    if(res < 0) {
        res += MOD;
    }
    return res;
}

int main() {
    int N;
    cin >> N;
    ll K;
    cin >> K;
    vector<int> vec(N);
    for(int i = 0;i < N;i++) {
        cin >> vec.at(i);
        vec.at(i)--;
    }
    if(K <= 100000) {
        int cnt2 = 0;
        for(int i = 0;i < K;i++) {
            cnt2 = vec.at(cnt2);
        }
        cout << cnt2 + 1 << endl;
        return 0;
    }
    vector<ll> cnt(N,-1);
    cnt.at(0) = 0;
    int cnt2 = 0;
    ll cnt3 = 1;
    ll cnt4 = 0;
    while(true) {
        cnt2 = vec.at(cnt2);
        if(cnt.at(cnt2) != -1) {
            cnt4 = cnt3 - cnt.at(cnt2);
            cnt3 = cnt.at(cnt2);
            break;
        }
        cnt.at(cnt2) = cnt3;
        cnt3++;
    }
    K -= cnt3;
    for(int i = 0;i < N;i++) {
        cnt.at(i) -= cnt3;
    }
    for(int i = 0;i < N;i++) {
        if(cnt.at(i) >= 0 && cnt.at(i) == K % cnt4) {
            cout << i + 1 << endl;
            break;
        }
    }
}