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;
  long long a, b, T, q, w, e, d[100001], v[100001], t[100001], B = 150;
  int main() {
    scanf("%lld%lld", &a, &b);
    for (int i = 1; i <= a; i++) scanf("%lld", &d[i]), v[d[i]] = i, t[i] = i;
    for (int i = 1; i <= a; i++)
      for (int j = 1; j <= B; j++) t[i] = d[t[i]];
    for (int ii = 1; ii <= b; ii++) {
      scanf("%lld%lld%lld", &q, &w, &e);
      if (q == 1) {
        swap(d[w], d[e]);
        swap(v[d[w]], v[d[e]]);
        long long tt = e, yy = e;
        for (int i = 1; i <= B; i++) yy = d[yy];
        for (int i = 1; i <= B + 1; i++) t[tt] = yy, tt = v[tt], yy = v[yy];
        tt = w, yy = w;
        for (int i = 1; i <= B; i++) yy = d[yy];
        for (int i = 1; i <= B + 1; i++) t[tt] = yy, tt = v[tt], yy = v[yy];
      } else {
        long long ff = w;
        for (int i = 1; i <= e / B; i++) ff = t[ff];
        e %= B;
        for (int i = 1; i <= e; i++) ff = d[ff];
        printf("%lld\n", ff);
      }
    }
    return 0;
  }
  