#include <bits/stdc++.h>
using namespace std;
long long ne[1000005], pre[1000005], ne300[1000005], pre300[1000005];
void change(long long x) {
  long long k = x;
  for (long long i = 1; i < 300; i++) k = pre[k];
  for (long long i = 1, k2 = ne[x]; i <= 300; i++) {
    ne300[k] = k2;
    pre300[k2] = k;
    k = ne[k];
    k2 = ne[k2];
  }
  return;
}
int main() {
  long long n, q;
  scanf("%lld%lld", &n, &q);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &ne[i]);
    pre[ne[i]] = i;
  }
  for (long long i = 1; i <= n; i++)
    for (long long j = 1, k = i; j <= 300; j++)
      k = ne[k], ne300[i] = k, pre300[k] = i;
  while (q--) {
    long long op, x, y;
    scanf("%lld%lld%lld", &op, &x, &y);
    if (op == 1) {
      swap(ne[x], ne[y]);
      pre[ne[y]] = y, pre[ne[x]] = x;
      change(x), change(y);
    } else {
      while (y >= 300) x = ne300[x], y -= 300;
      while (y--) x = ne[x];
      printf("%lld\n", x);
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
long long ne[1000005], pre[1000005], ne300[1000005], pre300[1000005];
void change(long long x) {
  long long k = x;
  for (long long i = 1; i < 300; i++) k = pre[k];
  for (long long i = 1, k2 = ne[x]; i <= 300; i++) {
    ne300[k] = k2;
    pre300[k2] = k;
    k = ne[k];
    k2 = ne[k2];
  }
  return;
}
int main() {
  long long n, q;
  scanf("%lld%lld", &n, &q);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &ne[i]);
    pre[ne[i]] = i;
  }
  for (long long i = 1; i <= n; i++)
    for (long long j = 1, k = i; j <= 300; j++)
      k = ne[k], ne300[i] = k, pre300[k] = i;
  while (q--) {
    long long op, x, y;
    scanf("%lld%lld%lld", &op, &x, &y);
    if (op == 1) {
      swap(ne[x], ne[y]);
      pre[ne[y]] = y, pre[ne[x]] = x;
      change(x), change(y);
    } else {
      while (y >= 300) x = ne300[x], y -= 300;
      while (y--) x = ne[x];
      printf("%lld\n", x);
    }
  }
  return 0;
}

```