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;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, q;
    cin >> n >> q;
    int JUMP = int(0.55 * sqrt(n)) + 1;
    vector<int> p(n + 1), p_rev(n + 1), p_jump(n + 1);
    auto compute_p_jump = [&](int u) -> int {
      int steps = JUMP;
      while (steps--) u = p[u];
      return u;
    };
    for (int i = 1; i <= n; i++) {
      cin >> p[i];
      p_rev[p[i]] = i;
    }
    for (int i = 1; i <= n; i++) {
      p_jump[i] = compute_p_jump(i);
    }
    while (q--) {
      int op;
      cin >> op;
      if (op == 1) {
        int x, y;
        cin >> x >> y;
        swap(p[x], p[y]);
        swap(p_rev[p[x]], p_rev[p[y]]);
        int px = compute_p_jump(x);
        int py = compute_p_jump(y);
        for (int i = 0; i < JUMP; i++) {
          p_jump[x] = px;
          p_jump[y] = py;
          x = p_rev[x];
          y = p_rev[y];
          px = p_rev[px];
          py = p_rev[py];
        }
      } else {
        int id, k;
        cin >> id >> k;
        while (k >= JUMP) {
          id = p_jump[id];
          k -= JUMP;
        }
        while (k--) {
          id = p[id];
        }
        cout << id << "\n";
      }
    }
    return 0;
  }
  