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;
  mt19937_64 mrand(chrono::steady_clock::now().time_since_epoch().count());
  const int N = 1e5 + 5;
  int ch[N][2], rd[N], sz[N], val[N], a[N], fa[N], vis[N], tot = 0;
  void pushup(int rt) { sz[rt] = 1 + sz[ch[rt][0]] + sz[ch[rt][1]]; }
  void split(int rt, int k, int &x, int &y) {
    if (!rt) {
      x = y = 0;
      return;
    }
    if (sz[ch[rt][0]] + 1 <= k) {
      if (ch[rt][1]) fa[ch[rt][1]] = 0;
      x = rt, split(ch[rt][1], k - sz[ch[rt][0]] - 1, ch[rt][1], y);
      if (ch[rt][1]) fa[ch[rt][1]] = rt;
    } else {
      if (ch[rt][0]) fa[ch[rt][0]] = 0;
      y = rt, split(ch[rt][0], k, x, ch[rt][0]);
      if (ch[rt][0]) fa[ch[rt][0]] = rt;
    }
    pushup(rt);
  }
  int merge(int x, int y) {
    if (!x || !y) return x + y;
    if (rd[x] < rd[y]) {
      ch[x][1] = merge(ch[x][1], y);
      if (ch[x][1]) fa[ch[x][1]] = x;
      pushup(x);
      return x;
    } else {
      ch[y][0] = merge(x, ch[y][0]);
      if (ch[y][0]) fa[ch[y][0]] = y;
      pushup(y);
      return y;
    }
  }
  pair<int, int> get(int x) {
    int p = x, pos = sz[ch[x][0]];
    while (fa[p]) {
      if (p == ch[fa[p]][1]) pos += sz[ch[fa[p]][0]] + 1;
      p = fa[p];
    }
    return {p, pos};
  }
  int kth(int p, int k) {
    while (k) {
      if (k <= sz[ch[p][0]]) {
        p = ch[p][0];
      } else if (k == sz[ch[p][0]] + 1) {
        return p;
      } else {
        k -= sz[ch[p][0]] + 1;
        p = ch[p][1];
      }
    }
    return -1;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, q;
    cin >> n >> q;
    for (int i = 1; i <= n; ++i)
      cin >> a[i], rd[i] = mrand() % INT_MAX, sz[i] = 1;
    for (int i = 1; i <= n; ++i) {
      if (vis[i]) continue;
      int p = i, rt = 0;
      while (!vis[p]) {
        vis[p] = 1;
        rt = merge(rt, p);
        p = a[p];
      }
    }
    while (q--) {
      int op;
      cin >> op;
      if (op == 1) {
        int x, y;
        cin >> x >> y;
        auto [rtx, px] = get(x);
        auto [rty, py] = get(y);
        if (rtx == rty) {
          if (px > py) {
            swap(px, py);
            swap(x, y);
          }
          int l, m, r;
          split(rtx, px + 1, l, r);
          split(r, py - px, m, r);
          merge(l, r);
        } else {
          int lx, rx, ly, ry;
          split(rtx, px + 1, lx, rx);
          split(rty, py + 1, ly, ry);
          merge(merge(lx, ry), merge(ly, rx));
        }
      } else {
        int x, k;
        cin >> x >> k;
        auto [rt, pos] = get(x);
        pos = (pos + k) % sz[rt];
        cout << kth(rt, pos + 1) << '\n';
      }
    }
  }
  