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;
  string _reset = "\u001b[0m", _yellow = "\u001b[33m", _bold = "\u001b[1m";
  void DBG() { cerr << "]" << _reset << endl; }
  template <class H, class... T>
  void DBG(H h, T... t) {
    cerr << to_string(h);
    if (sizeof...(t)) cerr << ", ";
    DBG(t...);
  }
  const int INF = 1e9 + 7, N = 1e5 + 7, K = 320;
  int p[N], jump[N], ip[N];
  bool vis[N];
  signed main() {
    std::ios::sync_with_stdio(0);
    std::cin.tie(0);
    ;
    int n, q;
    cin >> n >> q;
    for (int i = 0; i < n; i++) {
      cin >> p[i];
      p[i]--;
      ip[p[i]] = i;
    }
    for (int i = 0; i < n; i++) {
      jump[i] = i;
      for (int j = 0; j < K; j++) jump[i] = p[jump[i]];
    }
    auto fix = [&](vector<int>& v) {
      assert(v.size());
      jump[v[0]] = v[0];
      for (int j = 0; j < K; j++) jump[v[0]] = p[jump[v[0]]];
      for (int i = 1; i < int(v.size()); i++) jump[v[i]] = p[jump[v[i - 1]]];
    };
    while (q--) {
      int t, x, y;
      cin >> t >> x >> y;
      if (t == 1) {
        x--, y--;
        vector<int> fix1, fix2;
        {
          int u = x;
          fix1.push_back(u);
          while (ip[u] != x && ip[u] != y && int(fix1.size()) < K) {
            u = ip[u];
            fix1.push_back(u);
          }
          reverse((fix1).begin(), (fix1).end());
        }
        {
          int u = y;
          fix2.push_back(u);
          while (ip[u] != x && ip[u] != y && int(fix2.size()) < K) {
            u = ip[u];
            fix2.push_back(u);
          }
          reverse((fix2).begin(), (fix2).end());
        }
        swap(p[x], p[y]);
        swap(ip[p[x]], ip[p[y]]);
        fix(fix1), fix(fix2);
      } else {
        x--;
        while (y >= K) {
          x = jump[x];
          y -= K;
        }
        while (y) {
          x = p[x];
          y--;
        }
        cout << x + 1 << '\n';
      }
    }
  }
  