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;
  const int N = 1e5 + 10;
  int n, q;
  int pre[N], nxt[N];
  int sq;
  int num;
  int cnt;
  int now;
  int c[N], f[N], t[N];
  vector<int> need_resolve;
  int calc_nxt(int x) {
    int ret = x;
    for (int i = 1; i <= sq; ++i) {
      ret = nxt[ret];
    }
    return ret;
  }
  int calc_pre(int x) {
    int ret = x;
    for (int i = 1; i <= sq; ++i) {
      ret = pre[ret];
    }
    return ret;
  }
  int main() {
    scanf("%d%d", &n, &q);
    sq = sqrt(n);
    sq = min(sq, 150);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", nxt + i);
      pre[nxt[i]] = i;
    }
    for (int i = 1; i <= n; ++i) {
      f[i] = calc_nxt(i);
    }
    while (q--) {
      int op, x, y;
      scanf("%d%d%d", &op, &x, &y);
      int nx = nxt[x], ny = nxt[y];
      if (op == 1) {
        need_resolve.clear();
        need_resolve.emplace_back(calc_pre(x));
        need_resolve.emplace_back(calc_pre(y));
        need_resolve.emplace_back(calc_pre(nx));
        need_resolve.emplace_back(calc_pre(ny));
        nxt[x] = ny;
        pre[ny] = x;
        nxt[y] = nx;
        pre[nx] = y;
        need_resolve.emplace_back(calc_pre(nx));
        need_resolve.emplace_back(calc_pre(ny));
        for (auto u : need_resolve) {
          t[1] = u;
          for (int i = 2; i <= 2 * sq + 2; ++i) {
            t[i] = nxt[t[i - 1]];
          }
          for (int i = 1; i <= sq + 2; ++i) {
            f[t[i]] = t[i + sq];
          }
        }
      } else {
        int ret = x;
        for (; y > sq; y -= sq) {
          ret = f[ret];
        }
        for (; y; --y) {
          ret = nxt[ret];
        }
        printf("%d\n", ret);
      }
    }
  }
  