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;
  const int N = 1e5 + 10, magic = sqrt(N);
  int p[N], r[N], J[N];
  int jump(int i, int z = magic) {
    int k = i;
    for (int j = 1; j <= z; j++) {
      k = p[k];
      if (k == i) return -1;
    }
    return k;
  }
  int recalc(int x) {
    int y = jump(x);
    if (y < 0) return 0;
    for (int j = 1; j <= magic + 10; j++) {
      J[x] = y;
      x = r[x];
      y = r[y];
    }
    return 1;
  }
  void dfs_small(int v, int s, vector<int> &cycle) {
    if (v == s and cycle.size()) return;
    cycle.push_back(v);
    dfs_small(p[v], s, cycle);
  }
  int32_t main() {
    ios_base ::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, q;
    cin >> n >> q;
    for (int i = 1; i <= n; ++i) {
      cin >> p[i];
      r[p[i]] = i;
    }
    for (int i = 1; i <= n; i++) {
      J[i] = jump(i);
    }
    for (int i = 1; i <= q; i++) {
      int t;
      cin >> t;
      if (t == 1) {
        int x, y;
        cin >> x >> y;
        swap(r[p[x]], r[p[y]]);
        swap(p[x], p[y]);
        recalc(x);
        recalc(y);
      } else {
        int j, k;
        cin >> j >> k;
        if (jump(j) == -1) {
          vector<int> cycle;
          dfs_small(j, j, cycle);
          int len = cycle.size();
          cout << cycle[k % len] << endl;
        } else {
          while (k >= magic) {
            j = J[j];
            k -= magic;
          }
          cout << jump(j, k) << endl;
        }
      }
    }
  }
  