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 N, Q, K;
  int P[100005], R[100005], G[100005];
  void precalc() {
    for (int i = 1; i <= N; i++)
      if (!G[i]) {
        int b = i;
        for (int j = 0; j < K; j++) b = P[b];
        for (int j = i; !G[j]; j = P[j], b = P[b]) G[j] = b;
      }
  }
  int get(int n, int step) {
    while (step >= K) {
      n = G[n];
      step -= K;
    }
    while (step--) {
      n = P[n];
    }
    return n;
  }
  void refresh(int n) {
    int m = n;
    for (int i = 0; i < K; i++) m = P[m];
    G[n] = m;
    for (int i = 0; i < K; i++) {
      n = R[n];
      m = R[m];
      G[n] = m;
    }
  }
  void do_swap(int a, int b) {
    swap(P[a], P[b]);
    swap(R[P[a]], R[P[b]]);
    refresh(a);
    refresh(b);
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> N >> Q;
    K = sqrt(N);
    for (int i = 1; i <= N; i++) {
      cin >> P[i];
      R[P[i]] = i;
    }
    precalc();
    while (Q--) {
      int t, a, b;
      cin >> t >> a >> b;
      if (t == 2)
        cout << get(a, b) << '\n';
      else
        do_swap(a, b);
    }
  }
  