#include <bits/stdc++.h>
using namespace std;
long long a, b, T, q, w, e, d[100001], v[100001], t[100001], B = 300;
int main() {
  scanf("%lld%lld", &a, &b);
  for (int i = 1; i <= a; i++) scanf("%lld", &d[i]), v[d[i]] = i, t[i] = i;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= B; j++) t[i] = d[t[i]];
  for (int ii = 1; ii <= b; ii++) {
    scanf("%lld%lld%lld", &q, &w, &e);
    if (q == 1) {
      swap(d[w], d[e]);
      swap(v[d[w]], v[d[e]]);
      long long tt = e, yy = e;
      for (int i = 1; i <= B; i++) yy = d[yy];
      for (int i = 1; i <= B + 1; i++) t[tt] = yy, tt = v[tt], yy = v[yy];
      tt = w, yy = w;
      for (int i = 1; i <= B; i++) yy = d[yy];
      for (int i = 1; i <= B + 1; i++) t[tt] = yy, tt = v[tt], yy = v[yy];
    } else {
      long long ff = w;
      for (int i = 1; i <= e / B; i++) ff = t[ff];
      e %= B;
      for (int i = 1; i <= e; i++) ff = d[ff];
      printf("%lld\n", ff);
    }
  }
  return 0;
}

### 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;
long long a, b, T, q, w, e, d[100001], v[100001], t[100001], B = 300;
int main() {
  scanf("%lld%lld", &a, &b);
  for (int i = 1; i <= a; i++) scanf("%lld", &d[i]), v[d[i]] = i, t[i] = i;
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= B; j++) t[i] = d[t[i]];
  for (int ii = 1; ii <= b; ii++) {
    scanf("%lld%lld%lld", &q, &w, &e);
    if (q == 1) {
      swap(d[w], d[e]);
      swap(v[d[w]], v[d[e]]);
      long long tt = e, yy = e;
      for (int i = 1; i <= B; i++) yy = d[yy];
      for (int i = 1; i <= B + 1; i++) t[tt] = yy, tt = v[tt], yy = v[yy];
      tt = w, yy = w;
      for (int i = 1; i <= B; i++) yy = d[yy];
      for (int i = 1; i <= B + 1; i++) t[tt] = yy, tt = v[tt], yy = v[yy];
    } else {
      long long ff = w;
      for (int i = 1; i <= e / B; i++) ff = t[ff];
      e %= B;
      for (int i = 1; i <= e; i++) ff = d[ff];
      printf("%lld\n", ff);
    }
  }
  return 0;
}

```