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 JavaScript (Node.js 12.16.1) (judge verdict: Wrong Answer)

input = require("fs").readFileSync("/dev/stdin", "utf8").split("\n")

N = input[0].split(' ').map(i => parseInt(i))[0]
K = input[0].split(' ').map(i => parseInt(i))[1]

A = input[1].split(' ').map(i => parseInt(i))

// console.log(N, K, A)

before = []
after = []
pos = 0

for (i = 0; i <= K; i++) {
    // console.log(before, after, pos)
    if (after.length < 1) {
        if (!before.includes(A[pos])) {
            before.push(A[pos])
        } else {
            after.push(A[pos])
        }
    } else {
        before2 = new Array(...before)
        before3 = new Array(...before)
        // console.log('hoge', after, before2.splice(-after.length))
        if (JSON.stringify(after) == JSON.stringify(before3.splice(-after.length))) {
            console.log(A[i - after.length * 2 + K % after.length - 1])
            return
        } else {
            after.push(A[pos])
        }
        // console.log(K, i, K == i)
        if (K == i) {
            console.log(A[pos - 1])
            return
        }
    }
    pos = A[pos] - 1
}