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>
  #pragma GCC target("avx2")
  #pragma GCC optimize("Ofast")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  int way[100005][8];
  int mem[(1 << (8 + 2))];
  int p[100005];
  int rev[100005];
  void upd(int x) {
    int grv = (1 << (8 + 1));
    int tg = x;
    for (int i = 0; i <= (1 << (8 - 1)); i++) {
      mem[grv + i] = tg;
      tg = p[tg];
    }
    tg = x;
    for (int i = 0; i <= (1 << (8 - 1)); i++) {
      for (int j = 0; j < 8; j++) {
        way[tg][j] = mem[grv + (1 << j)];
      }
      tg = rev[tg];
      grv--;
      mem[grv] = tg;
    }
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, q;
    cin >> n >> q;
    for (int i = 0; i < n; i++) {
      cin >> p[i];
      p[i]--;
      rev[p[i]] = i;
    }
    for (int i = 0; i < n; i++) {
      int tg = i, np = 0;
      for (int j = 1; j <= (1 << (8 - 1)); j++) {
        tg = p[tg];
        if (j == (1 << np)) {
          way[i][np] = tg;
          np++;
        }
      }
    }
    while (q > 0) {
      q--;
      int t, x, y;
      cin >> t >> x >> y;
      if (t == 1) {
        x--;
        y--;
        swap(p[x], p[y]);
        rev[p[x]] = x;
        rev[p[y]] = y;
        upd(x);
        upd(y);
      } else {
        x--;
        int res = x;
        int cp = 8 - 1;
        while (y > 0) {
          while (y < (1 << cp)) {
            cp--;
          }
          y -= (1 << cp);
          res = way[res][cp];
        }
        cout << res + 1 << '\n';
      }
    }
    return 0;
  }
  