#include <bits/stdc++.h>
using namespace std;
int n, q, block, d[100005];
int p[100005], p2[100005];
int nxt(int x, int y = block) {
  while (y--) x = p[x];
  return x;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    p2[p[i]] = i;
  }
  block = sqrt(n);
  for (int i = 1; i <= n; i++) {
    d[i] = nxt(i, block);
  }
  while (q--) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      swap(p[x], p[y]);
      p2[p[x]] = x, p2[p[y]] = y;
      d[x] = nxt(x, block), d[y] = nxt(y, block);
      for (int i = 1; i <= block; i++) {
        d[p2[x]] = p2[d[x]];
        d[p2[y]] = p2[d[y]];
        x = p2[x], y = p2[y];
      }
    } else {
      while (y >= block) {
        x = d[x];
        y -= block;
      }
      while (y >= 1) {
        x = p[x];
        y--;
      }
      printf("%d\n", x);
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 n, q, block, d[100005];
int p[100005], p2[100005];
int nxt(int x, int y = block) {
  while (y--) x = p[x];
  return x;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    p2[p[i]] = i;
  }
  block = sqrt(n);
  for (int i = 1; i <= n; i++) {
    d[i] = nxt(i, block);
  }
  while (q--) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    if (op == 1) {
      swap(p[x], p[y]);
      p2[p[x]] = x, p2[p[y]] = y;
      d[x] = nxt(x, block), d[y] = nxt(y, block);
      for (int i = 1; i <= block; i++) {
        d[p2[x]] = p2[d[x]];
        d[p2[y]] = p2[d[y]];
        x = p2[x], y = p2[y];
      }
    } else {
      while (y >= block) {
        x = d[x];
        y -= block;
      }
      while (y >= 1) {
        x = p[x];
        y--;
      }
      printf("%d\n", x);
    }
  }
  return 0;
}

```