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 ne[1000005], pre[1000005], ne300[1000005], pre300[1000005];
  void change(long long x) {
    long long k = x;
    for (long long i = 1; i < 300; i++) k = pre[k];
    for (long long i = 1, k2 = ne[x]; i <= 300; i++) {
      ne300[k] = k2;
      pre300[k2] = k;
      k = ne[k];
      k2 = ne[k2];
    }
    return;
  }
  int main() {
    long long n, q;
    scanf("%lld%lld", &n, &q);
    for (long long i = 1; i <= n; i++) {
      scanf("%lld", &ne[i]);
      pre[ne[i]] = i;
    }
    for (long long i = 1; i <= n; i++)
      for (long long j = 1, k = i; j <= 300; j++)
        k = ne[k], ne300[i] = k, pre300[k] = i;
    while (q--) {
      long long op, x, y;
      scanf("%lld%lld%lld", &op, &x, &y);
      if (op == 1) {
        swap(ne[x], ne[y]);
        pre[ne[y]] = y, pre[ne[x]] = x;
        change(x), change(y);
      } else {
        while (y >= 300) x = ne300[x], y -= 300;
        while (y--) x = ne[x];
        printf("%lld\n", x);
      }
    }
    return 0;
  }
  