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 long long inf = 4e18;
  const long long N = 2e5 + 10;
  template <typename T>
  bool ckmin(T& a, T b) {
    return b < a && (a = b, true);
  }
  template <typename T>
  bool ckmax(T& a, T b) {
    return b > a && (a = b, true);
  }
  void testCase() {
    long long n, m;
    cin >> n >> m;
    vector<long long> p(n), inv_p(n), reach(n);
    long long S = 1;
    while (S * S <= n) S++;
    for (long long& i : p) cin >> i, i--;
    auto compute = [&](long long x) {
      long long i = x;
      for (long long r = 0; r < S; r++) {
        x = p[x];
      }
      return reach[i] = x;
    };
    for (long long i = 0; i < n; i++) {
      compute(i);
      inv_p[p[i]] = i;
    }
    for (long long i = 0; i < m; i++) {
      long long t;
      cin >> t;
      if (t == 1) {
        long long x, y;
        cin >> x >> y;
        x--, y--;
        swap(inv_p[p[x]], inv_p[p[y]]);
        swap(p[x], p[y]);
        long long sx = compute(x), sy = compute(y);
        for (long long r = 0; r < S; r++) {
          reach[x] = sx, reach[y] = sy;
          x = inv_p[x], y = inv_p[y];
          sx = inv_p[sx], sy = inv_p[sy];
        }
      } else {
        long long x, k;
        cin >> x >> k;
        x--;
        while (k >= S) {
          x = reach[x];
          k -= S;
        }
        while (k--) {
          x = p[x];
        }
        cout << x + 1 << '\n';
      }
    }
  }
  int32_t main() {
    cin.tie(0)->sync_with_stdio(0);
    long long t_c = 1;
    while (t_c--) testCase();
  }
  