#include <bits/stdc++.h>
using namespace std;
template <typename T_vector>
void output_vector(std::vector<T_vector>& v) {
  for (int i = 0; i < v.size(); i++)
    std::cout << v[i] << (i == v.size() - 1 ? '\n' : ' ');
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N, Q;
  cin >> N >> Q;
  vector<int> P(N);
  for (auto& p : P) {
    cin >> p;
    p--;
  }
  vector<int> rev_P(N);
  for (int i = 0; i < N; i++) {
    rev_P[P[i]] = i;
  }
  int S = int(sqrt(N));
  vector<int> s_jump(N);
  auto compute_s_jump = [&](int index) -> int {
    int x = index;
    for (int i = 0; i < S; i++) x = P[x];
    return s_jump[index] = x;
  };
  for (int i = 0; i < N; i++) compute_s_jump(i);
  for (int q = 0; q < Q; q++) {
    int type;
    cin >> type;
    if (type == 1) {
      int x, y;
      cin >> x >> y;
      x--;
      y--;
      swap(rev_P[P[x]], rev_P[P[y]]);
      swap(P[x], P[y]);
      int sx = compute_s_jump(x);
      int sy = compute_s_jump(y);
      for (int it = 0; it <= S; it++) {
        s_jump[x] = sx;
        s_jump[y] = sy;
        x = rev_P[x];
        y = rev_P[y];
        sx = rev_P[sx];
        sy = rev_P[sy];
      }
    } else {
      int x, k;
      cin >> x >> k;
      x--;
      while (k >= S) {
        x = s_jump[x];
        k -= S;
      }
      while (k >= 1) {
        x = P[x];
        k--;
      }
      cout << x + 1 << '\n';
    }
  }
}

### Prompt

Your task is to create a cpp solution 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;
template <typename T_vector>
void output_vector(std::vector<T_vector>& v) {
  for (int i = 0; i < v.size(); i++)
    std::cout << v[i] << (i == v.size() - 1 ? '\n' : ' ');
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N, Q;
  cin >> N >> Q;
  vector<int> P(N);
  for (auto& p : P) {
    cin >> p;
    p--;
  }
  vector<int> rev_P(N);
  for (int i = 0; i < N; i++) {
    rev_P[P[i]] = i;
  }
  int S = int(sqrt(N));
  vector<int> s_jump(N);
  auto compute_s_jump = [&](int index) -> int {
    int x = index;
    for (int i = 0; i < S; i++) x = P[x];
    return s_jump[index] = x;
  };
  for (int i = 0; i < N; i++) compute_s_jump(i);
  for (int q = 0; q < Q; q++) {
    int type;
    cin >> type;
    if (type == 1) {
      int x, y;
      cin >> x >> y;
      x--;
      y--;
      swap(rev_P[P[x]], rev_P[P[y]]);
      swap(P[x], P[y]);
      int sx = compute_s_jump(x);
      int sy = compute_s_jump(y);
      for (int it = 0; it <= S; it++) {
        s_jump[x] = sx;
        s_jump[y] = sy;
        x = rev_P[x];
        y = rev_P[y];
        sx = rev_P[sx];
        sy = rev_P[sy];
      }
    } else {
      int x, k;
      cin >> x >> k;
      x--;
      while (k >= S) {
        x = s_jump[x];
        k -= S;
      }
      while (k >= 1) {
        x = P[x];
        k--;
      }
      cout << x + 1 << '\n';
    }
  }
}

```