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;
  void main0();
  int main() {
    clock_t start, end;
    ios::sync_with_stdio(false);
    cin.tie(0);
    main0();
    return 0;
  }
  const int dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
  const int dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
  const int N = 2e5 + 5;
  const int M = 1e5;
  const int B = 300;
  const int INF = 0x3f3f3f3f;
  const long long INFF = 0x3f3f3f3f3f3f3f3f;
  const int mod = 998244353;
  const double eps = 1e-6;
  mt19937 rnd(
      (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
  template <typename T>
  bool chkmin(T &x, T y) {
    return y < x ? x = y, 1 : 0;
  }
  template <typename T>
  bool chkmax(T &x, T y) {
    return y > x ? x = y, 1 : 0;
  }
  int n, m;
  int a[N];
  int pre[N], nxt[N];
  int preB[N], nxtB[N];
  int jump(int type, int u, int k) {
    if (type)
      while (k >= B) u = nxtB[u], k -= B;
    while (k) u = nxt[u], k--;
    return u;
  }
  void update(int u) {
    int v = jump(0, u, B);
    for (int i = 0; i <= B; ++i, u = pre[u], v = pre[v]) {
      preB[v] = u;
      nxtB[u] = v;
    }
  }
  void main0() {
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
      cin >> nxt[i];
      pre[nxt[i]] = i;
    }
    for (int u = 1; u <= n; ++u) {
      int v = jump(0, u, B);
      preB[v] = u;
      nxtB[u] = v;
    }
    for (int i = 1; i <= m; ++i) {
      int op;
      cin >> op;
      if (op == 1) {
        int u, v;
        cin >> u >> v;
        pre[nxt[u]] = v;
        pre[nxt[v]] = u;
        swap(nxt[u], nxt[v]);
        update(u);
        update(v);
      }
      if (op == 2) {
        int u, k;
        cin >> u >> k;
        cout << jump(1, u, k) << endl;
      }
    }
  }
  