// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define F first

#define S second

#define pii pair<int, int>

#define eb emplace_back

#define all(v) v.begin(), v.end()

#define rep(i, n) for (int i = 0; i < (n); ++i)

#define rep3(i, l, n) for (int i = l; i < (n); ++i)

#define sz(v) (int)v.size()

const int inf = 1e9 + 7;

const ll INF = 1e18;

const int mod = 1000000007;

#define abs(x) (x >= 0 ? x : -(x))

#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())

#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())

template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }

template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }

template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }

template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }

template<typename T> T pow(T a, int b) { return b ? pow(a * a, b / 2) * (b % 2 ? a : 1) : 1; }

ll modpow(ll a, int b, int _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }

template<class T> ostream& operator<<(ostream& os, const vector<T>& vec) { for (auto &vi: vec) os << vi << " "; return os; }

template<class T, class U> ostream& operator<<(ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }

template<typename T >inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }

template<class T> inline void add(T &a, int b) { a += b; if (a >= mod) a -= mod; }







void solve();



int main() {

    ios::sync_with_stdio(false);

    cin.tie(0);

    cout.tie(0);

    cout << fixed << setprecision(10);



    int T;

    // cin >> T;

    T = 1;



    while (T--) {

        solve();

    }

}



// https://atcoder.jp/contests/abc167/submissions/13082002

// 間違いの検証

void solve() {

    int n;

    cin >> n;

    string k;

    cin >> k;



    const int N = 200005;

    vector<vector<int> > G(N);

    vector<int> seen(N);

    int pos = -1; // cycle 中に含まれる頂点 pos

    stack<int> hist; // 訪問履歴



    auto dfs = [&](auto&& dfs, int v, int p) -> void {

        seen[v] = 1;

        hist.push(v);

        for (auto nv : G[v]) {

            if (nv == p) continue; // ---------- 問題点を探す ----------

            if (seen[nv] == 2) continue;



            if (seen[nv] == 1) { // cycle を検出

                pos = nv;

                return;

            }

            dfs(dfs, nv, v);

            if (pos != -1) return;

        }

        hist.pop();

        seen[v] = 2;

    };



    rep(i, n) {

        int b;

        cin >> b;

        b--;

        G[i].eb(b);

    }



    ll x_ = stoll(k);

    if (x_ <= 10000000) { // ループに入る前に終わる

        int now = 0;

        rep(i, x_) {

            now = G[now][0];

        }

        cout << now + 1 << endl;

        return;

    }



    dfs(dfs, 0, -1);



    // cycle 復元

    vector<int> cycle;

    while (!hist.empty()) {

        int t = hist.top();

        cycle.eb(t);

        hist.pop();

        if (t == pos) break;

    }

    reverse(all(cycle));

    // for (auto e : cycle) cout << e << " "; cout << endl;



    // 本当は cycle の size が 2

    if (sz(cycle) == 0) { // ---------- あってる? ----------

      ll k_ = stoll(k);

      vector<int> v(n + 5);

      v[1] = G[0][0];

      rep3(i, 2, n + 5) v[i] = G[v[i - 1]][0];

      // cout << "v " << v << endl;

      rep(i, n + 5) {

        if (v[i] == v[i + 2]) {

          k_ %= 2;

          (k_ = k_ - i % 2 + 2) %= 2;

          if (k_ % 2 == 0) cout << v[i] + 1 << endl;

          else cout << v[i + 1] + 1 << endl;

          return;

        }

      }

    }



    int cnt = 0, now = 0;

    while (1) {

        if (now == cycle[0]) {

            break;

        }

        now = G[now][0];

        cnt++;

    }

    int k3 = 0;

    rep(i, sz(k)) {

        k3 *= 10;

        k3 += k[i] - '0';

        k3 %= sz(cycle);

    }

    k3 -= cnt; // 先に足すと, 10 倍するのが効いてしまう

    while (k3 < 0) k3 += sz(cycle);

    k3 %= sz(cycle);



    now = cycle[0];

    while (k3--) {

        now = G[now][0];

    }

    cout << now + 1 << endl;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

#define F first

#define S second

#define pii pair<int, int>

#define eb emplace_back

#define all(v) v.begin(), v.end()

#define rep(i, n) for (int i = 0; i < (n); ++i)

#define rep3(i, l, n) for (int i = l; i < (n); ++i)

#define sz(v) (int)v.size()

#define endl '\n'

const int inf = 1000000007;

const ll INF = 1e18;

// int mod = 998244353;

int mod = 1000000007;

#define abs(x) (x >= 0 ? x : -(x))

#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())

#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())

template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }

template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }

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

ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

template<typename T, typename U> T pow_(T a, U b) { return b ? pow_(a * a, b / 2) * (b % 2 ? a : 1) : 1; }

ll modpow(ll a, ll b, ll _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }

template<class T, class U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }

template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { rep(i, sz(vec)) { if (i) os << " "; os << vec[i]; } return os; }

template<typename T> inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }

template<class T, class T2> inline void add(T &a, T2 b) { a += b; if (a >= mod) a -= mod; }





void solve();



int main() {

  ios::sync_with_stdio(false);

  cin.tie(0);

  cout.tie(0);

  cout << fixed << setprecision(10);

  int T;

  T = 1;

  while (T--) solve();

}



// https://atcoder.jp/contests/abc167/submissions/13006376

void solve() {

  int n;

  ll k;

  cin >> n >> k;

  vector<int> a(n);

  cin >> a;



  vector<int> c(n + 1), ord(n + 1, -1); // すでに訪れてたら ord != -1, debuged n to n + 1

  c[0] = 1;

  ord[1] = 0; // debuged

  rep3(i, 1, n + 1) {

    c[i] = a[c[i - 1] - 1];

    if (ord[c[i]] != -1) { // debuged, i to c[i]

      int becy = ord[c[i]]; 

      int cyc = i - ord[c[i]];

      // cout << "b c " << becy << " " << cyc << endl;

      // cout << c << endl << ord << endl;

      if (k <= becy) cout << c[k] << endl;

      else {

        (k = k - becy) %= cyc;

        // cout << k << endl;

        cout << c[becy + k] << endl; // debuged, k to becy + k

      }

      return;

    }

    ord[c[i]] = i;

    // cout << i << " " << c[i] << endl;

  }

}


// Fastest known solution for this problem: