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;
  template <class T>
  void minn(T& a, T b) {
    if (a > b) a = b;
  }
  template <class T>
  void maxx(T& a, T b) {
    if (a < b) a = b;
  }
  void debug() { return; }
  template <class H, class... T>
  void debug(H a, T... b) {
    cerr << a;
    debug(b...);
  }
  void is() { return; }
  template <class H, class... T>
  void is(H& a, T&... b) {
    cin >> a;
    is(b...);
  }
  void os() { return; }
  template <class H, class... T>
  void os(H a, T... b) {
    cout << a;
    os(b...);
  }
  const int mod = 1e9 + 7;
  const int N = 2e5 + 5;
  int n, q;
  int nx[N], pr[N], nx100[N];
  void sw(int x, int y) {
    swap(nx[x], nx[y]);
    pr[nx[x]] = x;
    pr[nx[y]] = y;
    int k = x;
    for (int i = 1; i < 100; i++) k = pr[k];
    for (int i = 1; i <= 100; i++) {
      nx100[k] = nx[x];
      x = nx[x];
      k = nx[k];
    }
    k = y;
    for (int i = 1; i < 100; i++) k = pr[k];
    for (int i = 1; i <= 100; i++) {
      nx100[k] = nx[y];
      y = nx[y];
      k = nx[k];
    }
  }
  void solve() {
    is(n, q);
    for (int i = 1; i <= n; i++) {
      is(nx[i]);
      pr[nx[i]] = i;
    }
    for (int i = 1; i <= n; i++) {
      int k = i;
      for (int j = 1; j <= 100; j++) {
        k = nx[k];
      }
      nx100[i] = k;
    }
    while (q--) {
      int t, x, y;
      is(t, x, y);
      if (t == 2) {
        int k = x;
        while (y >= 100) {
          y -= 100;
          k = nx100[k];
        }
        while (y--) k = nx[k];
        os(k, '\n');
      } else {
        sw(x, y);
      }
    }
  }
  signed main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int T = 1;
    while (T--) {
      solve();
    }
  }
  