#include <bits/stdc++.h>
using namespace std;
const unsigned long long MAX = (1e5) + 10;
const unsigned long long mod = 998244353;
const unsigned long long oo = 1000000000000000002;
int ar[MAX];
int fth[MAX];
int ans[MAX];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  int Q = sqrt(n);
  for (int i = 1; i <= n; i++) {
    cin >> ar[i];
    fth[ar[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    int pos = i;
    for (int j = 0; j < Q; j++) {
      pos = ar[pos];
    }
    ans[i] = pos;
  }
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int a, b;
      cin >> a >> b;
      swap(ar[a], ar[b]);
      swap(ans[a], ans[b]);
      fth[ar[a]] = a;
      fth[ar[b]] = b;
      int pos = a;
      int pos1 = b;
      for (int j = 0; j < Q; j++) {
        pos = ar[pos];
        pos1 = ar[pos1];
      }
      ans[a] = pos;
      ans[b] = pos1;
      int p1 = a, p2 = ans[a];
      for (int i = 0; i < Q; i++) {
        ans[p1] = p2;
        p1 = fth[p1];
        p2 = fth[p2];
      }
      p1 = b, p2 = ans[b];
      for (int i = 0; i < Q; i++) {
        ans[p1] = p2;
        p1 = fth[p1];
        p2 = fth[p2];
      }
    } else {
      int a, k;
      cin >> a >> k;
      k--;
      int res = a;
      while (k >= Q) {
        k -= Q;
        res = ans[res];
      }
      for (int i = 0; i < k; i++) {
        res = ar[res];
      }
      cout << ar[res] << "\n";
    }
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 unsigned long long MAX = (1e5) + 10;
const unsigned long long mod = 998244353;
const unsigned long long oo = 1000000000000000002;
int ar[MAX];
int fth[MAX];
int ans[MAX];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  int Q = sqrt(n);
  for (int i = 1; i <= n; i++) {
    cin >> ar[i];
    fth[ar[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    int pos = i;
    for (int j = 0; j < Q; j++) {
      pos = ar[pos];
    }
    ans[i] = pos;
  }
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int a, b;
      cin >> a >> b;
      swap(ar[a], ar[b]);
      swap(ans[a], ans[b]);
      fth[ar[a]] = a;
      fth[ar[b]] = b;
      int pos = a;
      int pos1 = b;
      for (int j = 0; j < Q; j++) {
        pos = ar[pos];
        pos1 = ar[pos1];
      }
      ans[a] = pos;
      ans[b] = pos1;
      int p1 = a, p2 = ans[a];
      for (int i = 0; i < Q; i++) {
        ans[p1] = p2;
        p1 = fth[p1];
        p2 = fth[p2];
      }
      p1 = b, p2 = ans[b];
      for (int i = 0; i < Q; i++) {
        ans[p1] = p2;
        p1 = fth[p1];
        p2 = fth[p2];
      }
    } else {
      int a, k;
      cin >> a >> k;
      k--;
      int res = a;
      while (k >= Q) {
        k -= Q;
        res = ans[res];
      }
      for (int i = 0; i < k; i++) {
        res = ar[res];
      }
      cout << ar[res] << "\n";
    }
  }
  return 0;
}

```