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>
  template <class T>
  inline T read() {
    T r = 0, f = 0;
    char c;
    while (!isdigit(c = getchar())) f |= (c == '-');
    while (isdigit(c)) r = (r << 1) + (r << 3) + (c ^ 48), c = getchar();
    return f ? -r : r;
  }
  template <class T>
  inline void swap(T &a, T &b) {
    T c = a;
    a = b;
    b = c;
  }
  int n, m, B, p[101101], r[101101], jump[101101];
  inline void Jump(int x) {
    int X = x;
    for (int i = 1; i <= B; i++) X = p[X];
    for (int i = 1; i <= B; i++) jump[x] = X, x = r[x], X = r[X];
  }
  inline void Modify() {
    int x = read<int>();
    int y = read<int>();
    swap(r[p[x]], r[p[y]]);
    swap(p[x], p[y]);
    Jump(x), Jump(y);
  }
  inline void work() {
    int opt = read<int>();
    if (opt == 1) return Modify();
    int x = read<int>(), k = read<int>();
    while (k >= B) k -= B, x = jump[x];
    while (k) x = p[x], k--;
    printf("%d\n", x);
  }
  int main() {
    n = read<int>();
    m = read<int>();
    B = sqrt(n);
    for (int i = 1; i <= n; i++) p[i] = read<int>(), r[p[i]] = i;
    for (int i = 1; i <= n; i++) {
      jump[i] = i;
      for (int j = 1; j <= B; j++) jump[i] = p[jump[i]];
    }
    for (int t = 1; t <= m; t++) work();
    return 0;
  }
  