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;
  using ll = long long;
  using ull = unsigned long long;
  using ld = long double;
  template <typename T1, typename T2>
  inline void chkmin(T1& x, const T2& y) {
    if (y < x) x = y;
  }
  template <typename T1, typename T2>
  inline void chkmax(T1& x, const T2& y) {
    if (x < y) x = y;
  }
  const int N = 1e5 + 10;
  const int B = 100;
  struct Node {
    int next_node;
    int prev_node;
    int jump;
  };
  int n, q;
  Node nodes[N];
  bool used[N];
  int find_jump(int v) {
    for (int i = 0; i < B; ++i) {
      v = nodes[v].next_node;
    }
    return v;
  }
  void build() {
    for (int i = 0; i < n; ++i) {
      if (used[i]) continue;
      int v = i;
      int jump = find_jump(i);
      while (!used[v]) {
        used[v] = true;
        nodes[v].jump = jump;
        v = nodes[v].next_node;
        jump = nodes[jump].next_node;
      }
    }
  }
  void relax_jumps(int v) {
    int jump = v;
    for (int i = 0; i < B; ++i) {
      v = nodes[v].prev_node;
    }
    for (int i = 0; i <= B; ++i) {
      nodes[v].jump = jump;
      v = nodes[v].next_node;
      jump = nodes[jump].next_node;
    }
  }
  void upd(int i, int j) {
    int next_i = nodes[i].next_node;
    int next_j = nodes[j].next_node;
    nodes[i].next_node = next_j;
    nodes[j].next_node = next_i;
    nodes[next_i].prev_node = j;
    nodes[next_j].prev_node = i;
    relax_jumps(i);
    relax_jumps(j);
  }
  int getAns(int v, int k) {
    while (k) {
      if (k >= B) {
        v = nodes[v].jump;
        k -= B;
      } else {
        v = nodes[v].next_node;
        --k;
      }
    }
    return v;
  }
  signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cout.precision(20), cout.setf(ios::fixed);
    cin >> n >> q;
    for (int i = 0; i < n; ++i) {
      cin >> nodes[i].next_node;
      --nodes[i].next_node;
      nodes[nodes[i].next_node].prev_node = i;
    }
    build();
    while (q--) {
      int t;
      cin >> t;
      if (t == 1) {
        int i, j;
        cin >> i >> j;
        --i;
        --j;
        upd(i, j);
      } else if (t == 2) {
        int v, k;
        cin >> v >> k;
        --v;
        cout << getAns(v, k) + 1 << '\n';
      } else {
        assert(false);
      }
    }
    return 0;
  }
  