#include <bits/stdc++.h>
using namespace std;
int n, m, i, j, x, y, z, p[100001], q[100001], pre[100001], a[100001], b = 300;
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), pre[a[i]] = i;
  for (i = 1; i <= n; i++) {
    p[i] = i;
    for (j = 1; j <= b; j++) p[i] = a[p[i]];
  }
  while (m--) {
    scanf("%d%d%d", &x, &y, &z);
    if (x == 1) {
      swap(a[y], a[z]);
      pre[a[y]] = y;
      pre[a[z]] = z;
      q[0] = y;
      for (i = 1; i <= b; i++) q[i] = a[q[i - 1]];
      for (i = 0; i <= b; i++) p[y] = q[b - i], y = pre[y];
      q[0] = z;
      for (i = 1; i <= b; i++) q[i] = a[q[i - 1]];
      for (i = 0; i <= b; i++) p[z] = q[b - i], z = pre[z];
    }
    if (x == 2) {
      while (z >= b) z -= b, y = p[y];
      while (z) z--, y = a[y];
      printf("%d\n", y);
    }
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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, m, i, j, x, y, z, p[100001], q[100001], pre[100001], a[100001], b = 300;
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), pre[a[i]] = i;
  for (i = 1; i <= n; i++) {
    p[i] = i;
    for (j = 1; j <= b; j++) p[i] = a[p[i]];
  }
  while (m--) {
    scanf("%d%d%d", &x, &y, &z);
    if (x == 1) {
      swap(a[y], a[z]);
      pre[a[y]] = y;
      pre[a[z]] = z;
      q[0] = y;
      for (i = 1; i <= b; i++) q[i] = a[q[i - 1]];
      for (i = 0; i <= b; i++) p[y] = q[b - i], y = pre[y];
      q[0] = z;
      for (i = 1; i <= b; i++) q[i] = a[q[i - 1]];
      for (i = 0; i <= b; i++) p[z] = q[b - i], z = pre[z];
    }
    if (x == 2) {
      while (z >= b) z -= b, y = p[y];
      while (z) z--, y = a[y];
      printf("%d\n", y);
    }
  }
}

```