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>
  namespace cltstream {
  char cltin[1048576 + 1], *ih = cltin, *it = cltin;
  inline char gc() {
    if (ih == it) {
      it = (ih = cltin) + fread(cltin, 1, 1048576, stdin);
      if (ih == it) return EOF;
    }
    return *ih++;
  }
  char cltout[1048576 + 1], *oh = cltout, *ot = cltout + 1048576;
  inline void pc(char c) {
    if (oh == ot) {
      fwrite(cltout, 1, 1048576, stdout);
      oh = cltout;
    }
    *oh++ = c;
  }
  template <typename _tp>
  inline void read(_tp& x) {
    int sn = 1;
    char c = gc();
    for (; c != 45 && (c < 48 || c > 57) && c != EOF; c = gc())
      ;
    if (c == 45) sn = -1, c = gc();
    for (x = 0; c >= 48 && c <= 57 && c != EOF;
         x = (x << 3) + (x << 1) + (c ^ 48), c = gc())
      ;
    x *= sn;
  }
  template <typename _tp>
  inline void write(_tp x, char text = -1) {
    if (x < 0) pc(45), x = -x;
    if (!x)
      pc(48);
    else {
      int digit[22];
      for (digit[0] = 0; x; digit[++digit[0]] = x % 10, x /= 10)
        ;
      for (; digit[0]; pc(digit[digit[0]--] ^ 48))
        ;
    }
    if (text >= 0) pc(text);
  }
  }  // namespace cltstream
  int n, m, sq;
  int p[100000 + 1], r[100000 + 1];
  int f[100000 + 1];
  inline void work(register int x) {
    f[x] = x;
    for (register int i = 1; i <= sq; ++i) f[x] = p[f[x]];
    for (register int i = 1, u = r[x]; i <= sq && u != x; ++i, u = r[u])
      f[u] = r[f[p[u]]];
  }
  int main() {
    cltstream::read(n);
    cltstream::read(m);
    sq = sqrt(n);
    for (register int i = 1; i <= n; ++i) {
      cltstream::read(p[i]);
      r[p[i]] = i;
    }
    for (register int i = 1; i <= n; ++i) {
      f[i] = i;
      for (register int j = 1; j <= sq; ++j) f[i] = p[f[i]];
    }
    for (; m; --m) {
      int t, x, y, k;
      int res;
      cltstream::read(t);
      if (t == 1) {
        cltstream::read(x);
        cltstream::read(y);
        std::swap(p[x], p[y]);
        std::swap(r[p[x]], r[p[y]]);
        work(x);
        work(y);
      } else {
        cltstream::read(x);
        cltstream::read(k);
        res = x;
        for (register int i = 1; i <= k / sq; ++i) res = f[res];
        for (register int i = 1; i <= k % sq; ++i) res = p[res];
        cltstream::write(res, 10);
      }
    }
    fwrite(cltstream::cltout, 1, cltstream::oh - cltstream::cltout, stdout),
        cltstream::oh = cltstream::cltout;
    return 0;
  }
  