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;
  std::mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
  const int N = 2e5 + 2;
  unsigned int w[N];
  int c[N][2], v[N], s[N], f[N], p[N];
  int kth;
  bool ed[N];
  inline void pushup(const int x) { s[x] = s[c[x][0]] + s[c[x][1]] + 1; }
  void split(int now, int &x, int &y) {
    if (!now) return x = y = 0, void();
    if (kth <= s[c[now][0]])
      f[c[now][0]] = 0, split(c[y = now][0], x, c[now][0]), f[c[now][0]] = now;
    else
      kth -= s[c[now][0]] + 1, f[c[now][1]] = 0,
                               split(c[x = now][1], c[now][1], y),
                               f[c[now][1]] = now;
    pushup(now);
  }
  int merge(int x, int y) {
    if (!(x && y)) return x | y;
    if (w[x] < w[y]) {
      c[x][1] = merge(c[x][1], y);
      f[c[x][1]] = x;
      pushup(x);
      return x;
    } else {
      c[y][0] = merge(x, c[y][0]);
      f[c[y][0]] = y;
      pushup(y);
      return y;
    }
  }
  int getkth(int x) {
    int r = s[c[x][0]] + 1;
    while (f[x]) {
      if (c[f[x]][1] == x) r += s[c[f[x]][0]] + 1;
      x = f[x];
    }
    return r;
  }
  int findroot(int x) {
    while (f[x]) x = f[x];
    return x;
  }
  int make_lst_root(int x) {
    kth = getkth(x);
    int rt = findroot(x);
    if (!kth) return rt;
    int p;
    split(rt, p, rt);
    return merge(rt, p);
  }
  int main() {
    int n, q, i, j;
    cin >> n >> q;
    for (i = 1; i <= n; i++) w[i] = rnd();
    for (i = 1; i <= n; i++) cin >> p[i], s[i] = 1;
    for (i = 1; i <= n; i++)
      if (!ed[i] && p[i] != i) {
        int rt = i;
        for (ed[i] = 1, j = p[i]; !ed[j]; j = p[j]) ed[j] = 1, rt = merge(rt, j);
      }
    while (q--) {
      int op, x, y;
      cin >> op >> x >> y;
      if (op == 1) {
        if (x == y) continue;
        int rt1 = make_lst_root(x);
        int rt2 = findroot(y);
        if (rt1 == rt2) {
          kth = getkth(y);
          int p;
          split(rt1, rt1, p);
        } else {
          make_lst_root(y);
          merge(rt1, rt2);
        }
      } else {
        int rt = make_lst_root(x);
        y %= s[rt];
        if (y == 0) {
          cout << x << '\n';
          continue;
        }
        int q;
        kth = y;
        split(rt, rt, q);
        int p = rt;
        while (c[p][1]) p = c[p][1];
        cout << p << '\n';
        merge(rt, q);
      }
    }
  }
  