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 = 2e5 + 7;
  const int inf = 0x3f3f3f3f;
  const long long INF = 0x3f3f3f3f3f3f3f3f;
  const long long mod = 1e9 + 7;
  inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (isdigit(ch)) {
      x = x * 10 + ch - 48;
      ch = getchar();
    }
    return x * f;
  }
  int n, q;
  int l[N], r[N], ans_sqrt[N];
  int Q = 100;
  void solve() {
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &r[i]);
      l[r[i]] = i;
    }
    for (int i = 1; i <= n; i++) {
      int p = i;
      for (int j = 0; j < Q; j++) {
        p = r[p];
      }
      ans_sqrt[i] = p;
    }
    while (q--) {
      int t, x, y;
      scanf("%d%d%d", &t, &x, &y);
      if (t == 1) {
        swap(r[x], r[y]);
        swap(l[r[x]], l[r[y]]);
        vector<int> vv;
        int pp = x;
        for (int p = 0; p < Q; p++) {
          pp = r[pp];
          vv.push_back(pp);
        }
        while (!vv.empty()) {
          ans_sqrt[x] = vv.back();
          vv.pop_back();
          x = l[x];
        }
        pp = y;
        for (int p = 0; p < Q; p++) {
          pp = r[pp];
          vv.push_back(pp);
        }
        while (!vv.empty()) {
          ans_sqrt[y] = vv.back();
          vv.pop_back();
          y = l[y];
        }
      } else {
        int ans = x;
        while (y >= Q) {
          y -= Q;
          ans = ans_sqrt[ans];
        }
        while (y--) {
          ans = r[ans];
        }
        printf("%d\n", ans);
      }
    }
  }
  int main() {
    int T = 1;
    while (T--) {
      solve();
    }
    return 0;
  }
  