#include <bits/stdc++.h>
using namespace std;
const int k = 316;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  vector<int> p(n), b(n), jump(n);
  for (int i = 0; i < n; ++i) {
    cin >> p[i];
    b[--p[i]] = i;
  }
  for (int i = 0; i < n; ++i) {
    jump[i] = i;
    for (int j = 0; j < k; ++j) jump[i] = p[jump[i]];
  }
  for (int i = 0; i < q; ++i) {
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      x--;
      y--;
      swap(p[x], p[y]);
      b[p[x]] = x;
      b[p[y]] = y;
      swap(jump[x], jump[y]);
      int bx = x, by = y;
      for (int j = 0; j < k - 1; ++j) {
        bx = b[bx];
        by = b[by];
      }
      int fx = p[x], fy = p[y];
      for (int j = 0; j < k - 1; ++j) {
        jump[bx] = fx;
        jump[by] = fy;
        fx = p[fx];
        fy = p[fy];
        bx = p[bx];
        by = p[by];
      }
    } else {
      x--;
      while (k <= y) x = jump[x], y -= k;
      for (int j = 0; j < y; ++j) x = p[x];
      cout << x + 1 << "\n";
    }
  }
}

### Prompt

Create a solution in cpp for 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 k = 316;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  vector<int> p(n), b(n), jump(n);
  for (int i = 0; i < n; ++i) {
    cin >> p[i];
    b[--p[i]] = i;
  }
  for (int i = 0; i < n; ++i) {
    jump[i] = i;
    for (int j = 0; j < k; ++j) jump[i] = p[jump[i]];
  }
  for (int i = 0; i < q; ++i) {
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      x--;
      y--;
      swap(p[x], p[y]);
      b[p[x]] = x;
      b[p[y]] = y;
      swap(jump[x], jump[y]);
      int bx = x, by = y;
      for (int j = 0; j < k - 1; ++j) {
        bx = b[bx];
        by = b[by];
      }
      int fx = p[x], fy = p[y];
      for (int j = 0; j < k - 1; ++j) {
        jump[bx] = fx;
        jump[by] = fy;
        fx = p[fx];
        fy = p[fy];
        bx = p[bx];
        by = p[by];
      }
    } else {
      x--;
      while (k <= y) x = jump[x], y -= k;
      for (int j = 0; j < y; ++j) x = p[x];
      cout << x + 1 << "\n";
    }
  }
}

```