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_vector>
  void output_vector(std::vector<T_vector>& v) {
    for (int i = 0; i < v.size(); i++)
      std::cout << v[i] << (i == v.size() - 1 ? '\n' : ' ');
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int N, Q;
    cin >> N >> Q;
    vector<int> P(N);
    for (auto& p : P) {
      cin >> p;
      p--;
    }
    vector<int> rev_P(N);
    for (int i = 0; i < N; i++) {
      rev_P[P[i]] = i;
    }
    int S = int(sqrt(N));
    vector<int> s_jump(N);
    auto compute_s_jump = [&](int index) -> int {
      int x = index;
      for (int i = 0; i < S; i++) x = P[x];
      return s_jump[index] = x;
    };
    for (int i = 0; i < N; i++) compute_s_jump(i);
    for (int q = 0; q < Q; q++) {
      int type;
      cin >> type;
      if (type == 1) {
        int x, y;
        cin >> x >> y;
        x--;
        y--;
        swap(rev_P[P[x]], rev_P[P[y]]);
        swap(P[x], P[y]);
        int sx = compute_s_jump(x);
        int sy = compute_s_jump(y);
        for (int it = 0; it <= S; it++) {
          s_jump[x] = sx;
          s_jump[y] = sy;
          x = rev_P[x];
          y = rev_P[y];
          sx = rev_P[sx];
          sy = rev_P[sy];
        }
      } else {
        int x, k;
        cin >> x >> k;
        x--;
        while (k >= S) {
          x = s_jump[x];
          k -= S;
        }
        while (k >= 1) {
          x = P[x];
          k--;
        }
        cout << x + 1 << '\n';
      }
    }
  }
  