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 maxn = 1e6 + 50;
  template <typename A, typename B>
  istream& operator>>(istream& o, pair<A, B>& a) {
    return o >> a.first >> a.second;
  }
  template <typename A, typename B>
  ostream& operator<<(ostream& o, pair<A, B> a) {
    return o << '(' << a.first << ", " << a.second << ')';
  }
  namespace io {
  const long long SIZE = 1e7 + 10;
  char inbuff[SIZE];
  char *l, *r;
  inline void init() {
    l = inbuff;
    r = inbuff + fread(inbuff, 1, SIZE, stdin);
  }
  inline char gc() {
    if (l == r) init();
    return (l != r) ? *(l++) : EOF;
  }
  void R(long long& x) {
    bool neg = 0;
    x = 0;
    char ch = gc();
    while (!isdigit(ch)) {
      if (ch == '-') neg = 1;
      ch = gc();
    }
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = gc();
    if (neg) x = -x;
  }
  }  // namespace io
  using io::R;
  void solve() {
    long long n, q, sz = 300, tot = 0;
    R(n), R(q);
    vector<long long> a(n), id(n), jp(n), r(n);
    for (long long& o : a) R(o), o--;
    for (long long i = 0; i <= n - 1; i++) {
      if (!id[a[i]]) {
        tot++;
        long long t = a[i];
        while (!id[t]) {
          id[t] = tot;
          r[a[t]] = t;
          t = a[t];
        }
      }
    }
    for (long long i = 0; i <= n - 1; i++) {
      long long t = i;
      for (long long j = 1; j <= sz; j++) t = a[t];
      jp[i] = t;
    }
    while (q--) {
      long long o, x, y;
      R(o), R(x), R(y);
      x--;
      if (o == 1) {
        y--;
        long long tx = x, gpx = x, ty = y, gpy = y;
        r[a[x]] = y;
        r[a[y]] = x;
        swap(a[x], a[y]);
        for (long long i = 1; i <= sz; i++) gpx = a[gpx], gpy = a[gpy];
        jp[x] = gpx, jp[y] = gpy;
        for (long long i = 1; i <= sz; i++) {
          tx = r[tx];
          ty = r[ty];
          gpx = r[gpx];
          gpy = r[gpy];
          jp[tx] = gpx;
          jp[ty] = gpy;
        }
      } else {
        long long t = x;
        for (long long i = 1; i <= y / sz; i++) t = jp[t];
        for (long long i = 1; i <= y - y / sz * sz; i++) t = a[t];
        printf("%lld\n", t + 1);
      }
    }
  }
  void input() {}
  signed main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    ;
    long long tt = 1;
    while (tt--) {
      input();
      solve();
    }
  }
  