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;
  template <typename _T>
  void rd(_T &x) {
    int f = 1;
    x = 0;
    char s = getchar();
    while (s > '9' || s < '0') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s >= '0' && s <= '9')
      x = (x << 3) + (x << 1) + (s ^ 48), s = getchar();
    x *= f;
  }
  template <typename _T>
  void write(_T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
    return;
  }
  const int N = 100005, SQ = sqrt(N);
  int p[N], pinv[N], pa[N], nx[N];
  int main() {
    int n, q;
    rd(n), rd(q);
    for (register int i(1); i <= n; ++i) rd(p[i]), pinv[p[i]] = i;
    for (register int i(1); i <= n; ++i) {
      int a = i;
      for (register int j(1); j <= SQ; ++j) a = p[a];
      nx[i] = a;
    }
    while (q--) {
      int t, x, y, i, k;
      rd(t);
      if (t == 1) {
        rd(x), rd(y);
        swap(p[x], p[y]);
        pinv[p[x]] = x;
        pinv[p[y]] = y;
        int a = x;
        for (register int j(1); j <= SQ; ++j) a = p[a];
        nx[x] = a;
        for (register int i(1); i <= SQ; ++i) {
          x = pinv[x];
          a = pinv[a];
          nx[x] = a;
        }
        a = y;
        for (register int i(1); i <= SQ; ++i) a = p[a];
        nx[y] = a;
        for (register int i(1); i <= SQ; ++i) {
          y = pinv[y];
          a = pinv[a];
          nx[y] = a;
        }
      } else {
        rd(i), rd(k);
        while (k >= SQ) i = nx[i], k -= SQ;
        while (k--) i = p[i];
        printf("%d\n", i);
      }
    }
    return 0;
  }
  