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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.IOException;
import java.util.ArrayList;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) throws IOException {
//		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
//		String W	 = in.readLine();
//		int num = Integer.parseInt(W.split(" ")[0]);

		Scanner sc = new Scanner(System.in);
		long N = sc.nextLong();
		long[] A = new long[(int)N];

		long K = sc.nextLong();

		for(int i = 0;i<N ; i++) {
			A[i] = sc.nextLong();
		}
		boolean[] reachable = new boolean[(int) N];
		long[] reachedTime = new long[(int)N];
		long tmpTown = 1;
		long reachedNum=0;
		long roopminus=0;
		long roop = 0;


		for(int i = 0;i<N ; i++) {
			if(reachable[(int) (tmpTown-1)]) {
				roop = i- reachedTime[(int) (tmpTown-1)];
				roopminus = reachedTime[(int) (tmpTown-1)];
				break;
			}
			reachedNum++;
			reachable[(int) (tmpTown-1)]=true;
			reachedTime[(int) (tmpTown-1)] = i;
			tmpTown=A[(int) (tmpTown-1)];
		}


		K = (K-roopminus)%roop;

		long ans = 1;
		for(long i=0;i<roopminus;i++) {
			ans =  A[(int) (ans-1)];
		}

		for(long i=0;i<K;i++) {
			ans = A[(int) (ans-1)];
		}

		System.out.println(ans);

	}

	//	//二分探索
	//k <= num となる最小の配列要素kのインデックスを返す
	static private int binarySearch(long num, long[] orderedArray){
		int lowerBorder = -1;
		int upperBorder = orderedArray.length;
		int mid;

		while(upperBorder - lowerBorder >1) {
			mid = (upperBorder + lowerBorder)/2;
			if(orderedArray[mid]<=num) {
				lowerBorder = mid;
			}else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//二分探索
	//k <= num となる最小のList要素kのインデックスを返す
	static private int binarySearch(long num, ArrayList<Long> orderedList){
		int lowerBorder = -1;
		int upperBorder = orderedList.size();
		int mid;

		while(upperBorder - lowerBorder >1) {
			mid = (upperBorder + lowerBorder)/2;
			if(orderedList.get(mid)<=num) {
				lowerBorder = mid;
			}else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//aとbの最小公倍数を求める
	public static int gcd(int a, int b) {
        return b == 0 ? a: gcd(b, a % b);
	}
	public static long gcd(long a, long b) {
        return b == 0 ? a: gcd(b, a % b);
	}



}