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;
  inline int read() {
    int ans = 0, f = 1;
    char c = getchar();
    while (c > '9' || c < '0') {
      if (c == '-') f = -1;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      ans = (ans << 1) + (ans << 3) + c - '0';
      c = getchar();
    }
    return ans * f;
  }
  inline void write(int x) {
    if (x < 0) putchar('-'), x = -x;
    if (x / 10) write(x / 10);
    putchar((char)(x % 10) + '0');
  }
  template <typename T>
  inline T Abs(T a) {
    return a > 0 ? a : -a;
  };
  template <typename T, typename TT>
  inline T Min(T a, TT b) {
    return a < b ? a : b;
  }
  template <typename T, typename TT>
  inline T Max(T a, TT b) {
    return a < b ? b : a;
  }
  const int N = 1e5 + 5, BLOCK = 144;
  int n, q, up1[N], up2[N], dn[N];
  inline int climb(int x) {
    for (int i = 1; i <= BLOCK; ++i) x = up1[x];
    return x;
  }
  signed main() {
    n = read();
    q = read();
    for (int i = 1; i <= n; ++i) {
      up1[i] = read();
      dn[up1[i]] = i;
    }
    for (int i = 1; i <= n; ++i) up2[i] = climb(i);
    while (q--) {
      int opt = read();
      if (opt == 1) {
        int x = read(), y = read();
        swap(dn[up1[x]], dn[up1[y]]);
        swap(up1[x], up1[y]);
        int px = climb(x), py = climb(y);
        for (int i = 0; i <= BLOCK; ++i) {
          up2[x] = px, up2[y] = py;
          x = dn[x], px = dn[px];
          y = dn[y], py = dn[py];
        }
      } else {
        int x = read(), k = read();
        while (k >= BLOCK) x = up2[x], k -= BLOCK;
        while (k) x = up1[x], --k;
        write(x);
        puts("");
      }
    }
    return 0;
  }
  