#include <bits/stdc++.h>
using namespace std;
const int Q = 100;
int n, q, p[100005], r[100005], a[100005];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  for (int i = 0; i < n; i++) r[p[i]] = i;
  for (int i = 0; i < n; i++) {
    int first = i;
    for (int j = 0; j < Q; j++) first = p[first];
    a[i] = first;
  }
  while (q--) {
    int t, first, second;
    cin >> t >> first >> second;
    if (t == 2) {
      first--;
      while (second >= Q) {
        second -= Q;
        first = a[first];
      }
      while (second--) first = p[first];
      cout << first + 1 << "\n";
    } else {
      first--;
      second--;
      swap(r[p[first]], r[p[second]]);
      swap(p[first], p[second]);
      int ax = first;
      for (int i = 0; i < Q; i++) ax = p[ax];
      int x2 = first;
      for (int i = 0; i < Q; i++) {
        a[first] = ax;
        first = r[first];
        ax = r[ax];
      }
      swap(first, second);
      ax = first;
      for (int i = 0; i < Q; i++) ax = p[ax];
      x2 = first;
      for (int i = 0; i < Q; i++) {
        a[first] = ax;
        first = r[first];
        ax = r[ax];
      }
    }
  }
}

### Prompt

Please formulate a Cpp solution 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;
const int Q = 100;
int n, q, p[100005], r[100005], a[100005];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
  }
  for (int i = 0; i < n; i++) r[p[i]] = i;
  for (int i = 0; i < n; i++) {
    int first = i;
    for (int j = 0; j < Q; j++) first = p[first];
    a[i] = first;
  }
  while (q--) {
    int t, first, second;
    cin >> t >> first >> second;
    if (t == 2) {
      first--;
      while (second >= Q) {
        second -= Q;
        first = a[first];
      }
      while (second--) first = p[first];
      cout << first + 1 << "\n";
    } else {
      first--;
      second--;
      swap(r[p[first]], r[p[second]]);
      swap(p[first], p[second]);
      int ax = first;
      for (int i = 0; i < Q; i++) ax = p[ax];
      int x2 = first;
      for (int i = 0; i < Q; i++) {
        a[first] = ax;
        first = r[first];
        ax = r[ax];
      }
      swap(first, second);
      ax = first;
      for (int i = 0; i < Q; i++) ax = p[ax];
      x2 = first;
      for (int i = 0; i < Q; i++) {
        a[first] = ax;
        first = r[first];
        ax = r[ax];
      }
    }
  }
}

```