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>
  const int N = 1e5 + 5;
  const int block = 320;
  int p[N], pre[N], to[N];
  int n, q;
  inline int Read() {
    char ch;
    int f = 1;
    while ((ch = getchar()) < '0' || ch > '9')
      if (ch == '-') f = -1;
    int x = ch ^ 48;
    while ((ch = getchar()) >= '0' && ch <= '9')
      x = (x << 3) + (x << 1) + (ch ^ 48);
    return x * f;
  }
  inline void print(const int x) {
    if (x >= 10) print(x / 10);
    putchar(x % 10 + 48);
    return;
  }
  inline void Print(const int x, const char ch = '\n') {
    if (x < 0) {
      putchar('-');
      print(-x);
    } else
      print(x);
    putchar(ch);
    return;
  }
  inline void Update(const int u) {
    int v = u;
    for (register int i = 1; i <= block; i++) v = p[v];
    int uu = u;
    for (register int i = 1; i <= block; i++) {
      to[uu] = v;
      uu = pre[uu];
      v = pre[v];
    }
    return;
  }
  inline void Swap(int& x, int& y) {
    int tmp = x;
    x = y;
    y = tmp;
    return;
  }
  inline void Modify(const int u, const int v) {
    Swap(p[u], p[v]);
    pre[p[u]] = u;
    pre[p[v]] = v;
    Update(u);
    Update(v);
    return;
  }
  inline int Query(const int u, int k) {
    int uu = u;
    while (k >= block) {
      k -= block;
      uu = to[uu];
    }
    while (k) {
      k--;
      uu = p[uu];
    }
    return uu;
  }
  inline void Init() {
    n = Read();
    q = Read();
    for (register int i = 1; i <= n; i++) {
      p[i] = Read();
      pre[p[i]] = i;
    }
    for (register int i = 1; i <= n; i++) Update(i);
    return;
  }
  int main() {
    for (Init(); q; q--) {
      int opt = Read(), u = Read(), v = Read();
      if (opt == 1)
        Modify(u, v);
      else
        Print(Query(u, v));
    }
    return 0;
  }
  