description: You are given a permutation p of n elements. A permutation of n elements is an array of length n containing each integer from 1 to n exactly once. For example, [1, 2, 3] and [4, 3, 5, 1, 2] are permutations, but [1, 2, 4] and [4, 3, 2, 1, 2] are not permutations. You should perform q queries.
  
  There are two types of queries:
  
    * 1 x y — swap p_x and p_y. 
    * 2 i k — print the number that i will become if we assign i = p_i k times. 
  
  Input
  
  The first line contains two integers n and q (1 ≤ n, q ≤ 10^5).
  
  The second line contains n integers p_1, p_2, ..., p_n.
  
  Each of the next q lines contains three integers. The first integer is t (1 ≤ t ≤ 2) — type of query. If t = 1, then the next two integers are x and y (1 ≤ x, y ≤ n; x ≠ y) — first-type query. If t = 2, then the next two integers are i and k (1 ≤ i, k ≤ n) — second-type query.
  
  It is guaranteed that there is at least one second-type query.
  
  Output
  
  For every second-type query, print one integer in a new line — answer to this query.
  
  Examples
  
  Input
  
  
  5 4
  5 3 4 2 1
  2 3 1
  2 1 2
  1 1 3
  2 1 2
  
  
  Output
  
  
  4
  1
  2
  
  
  Input
  
  
  5 9
  2 3 5 1 4
  2 3 5
  2 5 5
  2 5 1
  2 5 3
  2 5 4
  1 5 4
  2 5 3
  2 2 5
  2 5 1
  
  
  Output
  
  
  3
  5
  4
  2
  3
  3
  3
  1
  
  Note
  
  In the first example p = \{5, 3, 4, 2, 1\}. 
  
  The first query is to print p_3. The answer is 4.
  
  The second query is to print p_{p_1}. The answer is 1.
  
  The third query is to swap p_1 and p_3. Now p = \{4, 3, 5, 2, 1\}.
  
  The fourth query is to print p_{p_1}. The answer is 2.
solution: #include <bits/stdc++.h>
  using namespace std;
  template <typename T>
  using min_heap = priority_queue<T, vector<T>, greater<T>>;
  template <typename T>
  using max_heap = priority_queue<T>;
  template <typename T>
  ostream &operator<<(ostream &os, min_heap<T> H) {
    while (!H.empty()) {
      os << H.top() << " ";
      H.pop();
    }
    os << "\n";
    return os << "";
  }
  template <typename T>
  ostream &operator<<(ostream &os, max_heap<T> H) {
    while (!H.empty()) {
      os << H.top() << " ";
      H.pop();
    }
    os << "\n";
    return os << "";
  }
  template <typename F, typename S>
  ostream &operator<<(ostream &os, pair<F, S> P) {
    return os << P.first << " " << P.second;
  }
  template <typename T>
  ostream &operator<<(ostream &os, vector<T> arr) {
    for (int i = 0; i < (int)arr.size(); i++) {
      os << arr[i] << " ";
    }
    return os << "";
  }
  template <typename T>
  ostream &operator<<(ostream &os, vector<vector<T>> matrix) {
    os << "\n";
    for (int i = 0; i < (int)matrix.size(); i++) {
      for (int j = 0; j < (int)matrix[i].size(); j++) {
        os << matrix[i][j] << " ";
      }
      os << "\n";
    }
    return os << "";
  }
  template <typename T>
  ostream &operator<<(ostream &os, set<T> S) {
    for (auto s : S) {
      os << s << " ";
    }
    return os << "";
  }
  template <typename T>
  ostream &operator<<(ostream &os, multiset<T> S) {
    for (auto s : S) {
      os << s << " ";
    }
    return os << "";
  }
  template <typename F, typename S>
  ostream &operator<<(ostream &os, map<F, S> M) {
    os << "\n";
    for (auto m : M) {
      os << m << "\n";
    }
    return os << "";
  }
  template <typename F, typename S>
  ostream &operator<<(ostream &os, multimap<F, S> M) {
    os << "\n";
    for (auto m : M) {
      os << m << "\n";
    }
    return os << "";
  }
  void solve() {
    int n, q;
    cin >> n >> q;
    int jr = (int)floor(sqrt(n)) + 1;
    vector<int> nxt(n + 1);
    vector<int> prev(n + 1);
    for (int i = 1; i <= n; i++) {
      cin >> nxt[i];
      prev[nxt[i]] = i;
    }
    vector<int> jump(n + 1);
    for (int i = 1; i <= (int)n; i++) {
      jump[i] = i;
      for (int j = 1; j <= (int)jr; j++) {
        jump[i] = nxt[jump[i]];
      }
    }
    for (int query = 0; query < (int)q; query++) {
      int type;
      cin >> type;
      if (type == 1) {
        int u, v;
        cin >> u >> v;
        swap(nxt[u], nxt[v]);
        prev[nxt[u]] = u;
        prev[nxt[v]] = v;
        int ind = u;
        vector<int> temp(jr + 1);
        temp[0] = ind;
        for (int i = 1; i <= (int)jr; i++) {
          temp[i] = nxt[temp[i - 1]];
        }
        int cur = ind;
        for (int i = 0; i <= (int)jr; i++) {
          jump[cur] = temp[jr - i];
          cur = prev[cur];
        }
        ind = v;
        temp[0] = ind;
        for (int i = 1; i <= (int)jr; i++) {
          temp[i] = nxt[temp[i - 1]];
        }
        cur = ind;
        for (int i = 0; i <= (int)jr; i++) {
          jump[cur] = temp[jr - i];
          cur = prev[cur];
        }
      } else {
        int x, k;
        cin >> x >> k;
        int ans = x;
        while (k > jr) {
          ans = jump[ans];
          k -= jr;
        }
        for (int i = 0; i < (int)k; i++) {
          ans = nxt[ans];
        }
        cout << ans << "\n";
      }
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int no_of_testcases = 1;
    for (int i = 1; i <= no_of_testcases; i++) {
      solve();
    }
    return 0;
  }
  