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;
  void solution() {
    constexpr long long S = 220;
    long long n, q;
    cin >> n >> q;
    vector<long long> p(n + 1);
    vector<long long> rev(n + 1);
    for (long long i = 1; i <= n; ++i) {
      cin >> p[i];
      rev[p[i]] = i;
    }
    vector<long long> next(n + 1);
    for (long long i = 1; i <= n; ++i) {
      long long x = i;
      for (long long j = 0; j < S; ++j) x = p[x];
      next[i] = x;
    }
    while (q--) {
      long long t;
      cin >> t;
      if (t == 1) {
        long long x, y;
        cin >> x >> y;
        swap(rev[p[x]], rev[p[y]]);
        swap(p[x], p[y]);
        long long sx = x;
        for (long long i = 0; i < S; ++i) {
          sx = p[sx];
        }
        long long sy = y;
        for (long long i = 0; i < S; ++i) {
          sy = p[sy];
        }
        for (long long i = 0; i <= S; ++i) {
          next[x] = sx, next[y] = sy;
          x = rev[x], y = rev[y];
          sx = rev[sx], sy = rev[sy];
        }
      } else {
        long long i, k;
        cin >> i >> k;
        while (k >= S) {
          i = next[i];
          k -= S;
        }
        while (k >= 1) {
          i = p[i];
          k -= 1;
        }
        cout << i << "\n";
      }
    }
  }
  int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    solution();
  }
  