#include <bits/stdc++.h>
using namespace std;
template <typename T>
using min_heap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using max_heap = priority_queue<T>;
template <typename T>
ostream &operator<<(ostream &os, min_heap<T> H) {
  while (!H.empty()) {
    os << H.top() << " ";
    H.pop();
  }
  os << "\n";
  return os << "";
}
template <typename T>
ostream &operator<<(ostream &os, max_heap<T> H) {
  while (!H.empty()) {
    os << H.top() << " ";
    H.pop();
  }
  os << "\n";
  return os << "";
}
template <typename F, typename S>
ostream &operator<<(ostream &os, pair<F, S> P) {
  return os << P.first << " " << P.second;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> arr) {
  for (int i = 0; i < (int)arr.size(); i++) {
    os << arr[i] << " ";
  }
  return os << "";
}
template <typename T>
ostream &operator<<(ostream &os, vector<vector<T>> matrix) {
  os << "\n";
  for (int i = 0; i < (int)matrix.size(); i++) {
    for (int j = 0; j < (int)matrix[i].size(); j++) {
      os << matrix[i][j] << " ";
    }
    os << "\n";
  }
  return os << "";
}
template <typename T>
ostream &operator<<(ostream &os, set<T> S) {
  for (auto s : S) {
    os << s << " ";
  }
  return os << "";
}
template <typename T>
ostream &operator<<(ostream &os, multiset<T> S) {
  for (auto s : S) {
    os << s << " ";
  }
  return os << "";
}
template <typename F, typename S>
ostream &operator<<(ostream &os, map<F, S> M) {
  os << "\n";
  for (auto m : M) {
    os << m << "\n";
  }
  return os << "";
}
template <typename F, typename S>
ostream &operator<<(ostream &os, multimap<F, S> M) {
  os << "\n";
  for (auto m : M) {
    os << m << "\n";
  }
  return os << "";
}
void solve() {
  int n, q;
  cin >> n >> q;
  int jr = (int)floor(sqrt(n)) + 1;
  vector<int> nxt(n + 1);
  vector<int> prev(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> nxt[i];
    prev[nxt[i]] = i;
  }
  vector<int> jump(n + 1);
  for (int i = 1; i <= (int)n; i++) {
    jump[i] = i;
    for (int j = 1; j <= (int)jr; j++) {
      jump[i] = nxt[jump[i]];
    }
  }
  for (int query = 0; query < (int)q; query++) {
    int type;
    cin >> type;
    if (type == 1) {
      int u, v;
      cin >> u >> v;
      swap(nxt[u], nxt[v]);
      prev[nxt[u]] = u;
      prev[nxt[v]] = v;
      int ind = u;
      vector<int> temp(jr + 1);
      temp[0] = ind;
      for (int i = 1; i <= (int)jr; i++) {
        temp[i] = nxt[temp[i - 1]];
      }
      int cur = ind;
      for (int i = 0; i <= (int)jr; i++) {
        jump[cur] = temp[jr - i];
        cur = prev[cur];
      }
      ind = v;
      temp[0] = ind;
      for (int i = 1; i <= (int)jr; i++) {
        temp[i] = nxt[temp[i - 1]];
      }
      cur = ind;
      for (int i = 0; i <= (int)jr; i++) {
        jump[cur] = temp[jr - i];
        cur = prev[cur];
      }
    } else {
      int x, k;
      cin >> x >> k;
      int ans = x;
      while (k > jr) {
        ans = jump[ans];
        k -= jr;
      }
      for (int i = 0; i < (int)k; i++) {
        ans = nxt[ans];
      }
      cout << ans << "\n";
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int no_of_testcases = 1;
  for (int i = 1; i <= no_of_testcases; i++) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
using min_heap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using max_heap = priority_queue<T>;
template <typename T>
ostream &operator<<(ostream &os, min_heap<T> H) {
  while (!H.empty()) {
    os << H.top() << " ";
    H.pop();
  }
  os << "\n";
  return os << "";
}
template <typename T>
ostream &operator<<(ostream &os, max_heap<T> H) {
  while (!H.empty()) {
    os << H.top() << " ";
    H.pop();
  }
  os << "\n";
  return os << "";
}
template <typename F, typename S>
ostream &operator<<(ostream &os, pair<F, S> P) {
  return os << P.first << " " << P.second;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> arr) {
  for (int i = 0; i < (int)arr.size(); i++) {
    os << arr[i] << " ";
  }
  return os << "";
}
template <typename T>
ostream &operator<<(ostream &os, vector<vector<T>> matrix) {
  os << "\n";
  for (int i = 0; i < (int)matrix.size(); i++) {
    for (int j = 0; j < (int)matrix[i].size(); j++) {
      os << matrix[i][j] << " ";
    }
    os << "\n";
  }
  return os << "";
}
template <typename T>
ostream &operator<<(ostream &os, set<T> S) {
  for (auto s : S) {
    os << s << " ";
  }
  return os << "";
}
template <typename T>
ostream &operator<<(ostream &os, multiset<T> S) {
  for (auto s : S) {
    os << s << " ";
  }
  return os << "";
}
template <typename F, typename S>
ostream &operator<<(ostream &os, map<F, S> M) {
  os << "\n";
  for (auto m : M) {
    os << m << "\n";
  }
  return os << "";
}
template <typename F, typename S>
ostream &operator<<(ostream &os, multimap<F, S> M) {
  os << "\n";
  for (auto m : M) {
    os << m << "\n";
  }
  return os << "";
}
void solve() {
  int n, q;
  cin >> n >> q;
  int jr = (int)floor(sqrt(n)) + 1;
  vector<int> nxt(n + 1);
  vector<int> prev(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> nxt[i];
    prev[nxt[i]] = i;
  }
  vector<int> jump(n + 1);
  for (int i = 1; i <= (int)n; i++) {
    jump[i] = i;
    for (int j = 1; j <= (int)jr; j++) {
      jump[i] = nxt[jump[i]];
    }
  }
  for (int query = 0; query < (int)q; query++) {
    int type;
    cin >> type;
    if (type == 1) {
      int u, v;
      cin >> u >> v;
      swap(nxt[u], nxt[v]);
      prev[nxt[u]] = u;
      prev[nxt[v]] = v;
      int ind = u;
      vector<int> temp(jr + 1);
      temp[0] = ind;
      for (int i = 1; i <= (int)jr; i++) {
        temp[i] = nxt[temp[i - 1]];
      }
      int cur = ind;
      for (int i = 0; i <= (int)jr; i++) {
        jump[cur] = temp[jr - i];
        cur = prev[cur];
      }
      ind = v;
      temp[0] = ind;
      for (int i = 1; i <= (int)jr; i++) {
        temp[i] = nxt[temp[i - 1]];
      }
      cur = ind;
      for (int i = 0; i <= (int)jr; i++) {
        jump[cur] = temp[jr - i];
        cur = prev[cur];
      }
    } else {
      int x, k;
      cin >> x >> k;
      int ans = x;
      while (k > jr) {
        ans = jump[ans];
        k -= jr;
      }
      for (int i = 0; i < (int)k; i++) {
        ans = nxt[ans];
      }
      cout << ans << "\n";
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int no_of_testcases = 1;
  for (int i = 1; i <= no_of_testcases; i++) {
    solve();
  }
  return 0;
}

```