#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, q;
  cin >> n >> q;
  int JUMP = int(0.43 * sqrt(n)) + 1;
  vector<int> p(n + 1), p_rev(n + 1), p_jump(n + 1);
  auto compute_p_jump = [&](int u) -> int {
    int steps = JUMP;
    while (steps--) u = p[u];
    return u;
  };
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    p_rev[p[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    p_jump[i] = compute_p_jump(i);
  }
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int x, y;
      cin >> x >> y;
      swap(p[x], p[y]);
      swap(p_rev[p[x]], p_rev[p[y]]);
      int px = compute_p_jump(x);
      int py = compute_p_jump(y);
      for (int i = 0; i < JUMP; i++) {
        p_jump[x] = px;
        p_jump[y] = py;
        x = p_rev[x];
        y = p_rev[y];
        px = p_rev[px];
        py = p_rev[py];
      }
    } else {
      int id, k;
      cin >> id >> k;
      while (k >= JUMP) {
        id = p_jump[id];
        k -= JUMP;
      }
      while (k--) {
        id = p[id];
      }
      cout << id << "\n";
    }
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, q;
  cin >> n >> q;
  int JUMP = int(0.43 * sqrt(n)) + 1;
  vector<int> p(n + 1), p_rev(n + 1), p_jump(n + 1);
  auto compute_p_jump = [&](int u) -> int {
    int steps = JUMP;
    while (steps--) u = p[u];
    return u;
  };
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    p_rev[p[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    p_jump[i] = compute_p_jump(i);
  }
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int x, y;
      cin >> x >> y;
      swap(p[x], p[y]);
      swap(p_rev[p[x]], p_rev[p[y]]);
      int px = compute_p_jump(x);
      int py = compute_p_jump(y);
      for (int i = 0; i < JUMP; i++) {
        p_jump[x] = px;
        p_jump[y] = py;
        x = p_rev[x];
        y = p_rev[y];
        px = p_rev[px];
        py = p_rev[py];
      }
    } else {
      int id, k;
      cin >> id >> k;
      while (k >= JUMP) {
        id = p_jump[id];
        k -= JUMP;
      }
      while (k--) {
        id = p[id];
      }
      cout << id << "\n";
    }
  }
  return 0;
}

```