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>
  #pragma GCC optimize("O3")
  using namespace std;
  using ll = long long;
  using P = pair<ll, ll>;
  template <class T>
  using V = vector<T>;
  const ll inf = (1e18);
  const ll mod = 1000000007;
  const vector<int> dy = {-1, 0, 1, 0}, dx = {0, -1, 0, 1};
  ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
  ll LCM(ll c, ll d) { return c / GCD(c, d) * d; }
  struct __INIT {
    __INIT() {
      cin.tie(0);
      ios::sync_with_stdio(false);
      cout << fixed << setprecision(15);
    }
  } __init;
  template <class T>
  bool chmax(T &a, const T &b) {
    if (a < b) {
      a = b;
      return 1;
    }
    return 0;
  }
  template <class T>
  bool chmin(T &a, const T &b) {
    if (a > b) {
      a = b;
      return 1;
    }
    return 0;
  }
  template <class T>
  void debag(const vector<T> &a) {
    cerr << "debag :";
    for (auto v : a) cerr << v << " ";
    cerr << "\n";
  }
  template <class T>
  void print(const vector<T> &a) {
    for (auto v : a) cout << v << " ";
    cout << "\n";
  }
  void solve() {
    int n, q;
    cin >> n >> q;
    int B = ll(sqrt(double(n))) + 10;
    V<int> f(n), fb(n), inv(n);
    iota((fb).begin(), (fb).end(), 0);
    for (int i = 0; i < n; i++) {
      cin >> f[i];
      f[i]--;
    }
    for (int i = 0; i < n; i++) {
      inv[f[i]] = i;
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < B; j++) {
        fb[i] = f[fb[i]];
      }
    }
    V<bool> used(n, true);
    for (int i = 0; i < q; i++) {
      int t, x, y;
      cin >> t >> x >> y;
      --x;
      --y;
      if (t == 1) {
        V<int> tmp;
        int cp = x;
        for (int j = 0; j <= B; j++) {
          tmp.emplace_back(cp);
          used[cp] = false;
          cp = inv[cp];
          if (cp == x) break;
        }
        cp = y;
        for (int j = 0; j <= B; j++) {
          tmp.emplace_back(cp);
          used[cp] = false;
          cp = inv[cp];
          if (cp == y) break;
        }
        swap(f[x], f[y]);
        swap(inv[f[x]], inv[f[y]]);
        for (int v : tmp) {
          if (!used[v]) {
            if (used[f[v]]) {
              fb[v] = inv[fb[f[v]]];
            } else {
              fb[v] = v;
              for (int j = 0; j < B; j++) {
                fb[v] = f[fb[v]];
              }
            }
            used[v] = true;
          }
        }
      } else {
        y++;
        int cur = x;
        while (B <= y) {
          cur = fb[cur];
          y -= B;
        }
        while (y > 0) {
          cur = f[cur];
          y--;
        }
        cout << cur + 1 << "\n";
      }
    }
  }
  int main() {
    int t = 1;
    while (t--) solve();
  }
  