#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
const int m = 3e2;
int crr = 0;
int t, n, q, type, x, y;
int L[maxn];
int R[maxn];
int upbl[maxn];
void update(int x) {
  int lhead = x;
  for (int i = 0; i < m; i++) x = R[x];
  for (int i = 0; i < m; i++) {
    upbl[lhead] = x;
    x = L[x];
    lhead = L[lhead];
  }
}
int main() {
  t = 1;
  while (t--) {
    cin >> n >> q;
    for (int i = 1; i <= n; i++) {
      cin >> R[i];
      upbl[i] = 0;
      L[R[i]] = i;
    }
    for (int i = 1; i <= n; i++)
      if (!upbl[i]) update(i);
    for (int i = 0; i < q; i++) {
      cin >> type >> x >> y;
      if (type == 1) {
        swap(R[x], R[y]);
        L[R[x]] = x;
        L[R[y]] = y;
        update(x);
        update(y);
      } else {
        while (y >= m) {
          x = upbl[x];
          y -= m;
        }
        while (y--) x = R[x];
        cout << x << "\n";
      }
    }
  }
}

### 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;
const int maxn = 2e5 + 5;
const int m = 3e2;
int crr = 0;
int t, n, q, type, x, y;
int L[maxn];
int R[maxn];
int upbl[maxn];
void update(int x) {
  int lhead = x;
  for (int i = 0; i < m; i++) x = R[x];
  for (int i = 0; i < m; i++) {
    upbl[lhead] = x;
    x = L[x];
    lhead = L[lhead];
  }
}
int main() {
  t = 1;
  while (t--) {
    cin >> n >> q;
    for (int i = 1; i <= n; i++) {
      cin >> R[i];
      upbl[i] = 0;
      L[R[i]] = i;
    }
    for (int i = 1; i <= n; i++)
      if (!upbl[i]) update(i);
    for (int i = 0; i < q; i++) {
      cin >> type >> x >> y;
      if (type == 1) {
        swap(R[x], R[y]);
        L[R[x]] = x;
        L[R[y]] = y;
        update(x);
        update(y);
      } else {
        while (y >= m) {
          x = upbl[x];
          y -= m;
        }
        while (y--) x = R[x];
        cout << x << "\n";
      }
    }
  }
}

```