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 i, j, k, n, t, x, y, z;
  int l[100500], r[100500], l300[100500], r300[100500], k2;
  void work(int x) {
    for (i = 1, k = x; i < 150; i++) k = l[k];
    for (i = 1, k2 = r[x]; i <= 150; i++) {
      r300[k] = k2;
      l300[k2] = k;
      k = r[k];
      k2 = r[k2];
    }
  }
  int main() {
    ios::sync_with_stdio(0);
    cin >> n >> t;
    for (i = 1; i <= n; i++) cin >> r[i], l[r[i]] = i;
    for (i = 1; i <= n; i++) {
      for (j = 1, k = i; j <= 150; j++) k = r[k];
      r300[i] = k;
      l300[k] = i;
    }
    while (t--) {
      cin >> x >> y >> z;
      if (x == 1) {
        swap(r[y], r[z]);
        l[r[y]] = y;
        l[r[z]] = z;
        work(y);
        work(z);
      } else {
        while (z > 150) y = r300[y], z -= 150;
        while (z--) y = r[y];
        cout << y << '\n';
      }
    }
  }
  