#include <bits/stdc++.h>
using namespace std;
int N, Q, K;
int P[100005], R[100005], G[100005];
void precalc() {
  for (int i = 1; i <= N; i++)
    if (!G[i]) {
      int b = i;
      for (int j = 0; j < K; j++) b = P[b];
      for (int j = i; !G[j]; j = P[j], b = P[b]) G[j] = b;
    }
}
int get(int n, int step) {
  while (step >= K) {
    n = G[n];
    step -= K;
  }
  while (step--) {
    n = P[n];
  }
  return n;
}
void refresh(int n) {
  int m = n;
  for (int i = 0; i < K; i++) m = P[m];
  G[n] = m;
  for (int i = 0; i < K; i++) {
    n = R[n];
    m = R[m];
    G[n] = m;
  }
}
void do_swap(int a, int b) {
  swap(P[a], P[b]);
  swap(R[P[a]], R[P[b]]);
  refresh(a);
  refresh(b);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> Q;
  K = sqrt(N);
  for (int i = 1; i <= N; i++) {
    cin >> P[i];
    R[P[i]] = i;
  }
  precalc();
  while (Q--) {
    int t, a, b;
    cin >> t >> a >> b;
    if (t == 2)
      cout << get(a, b) << '\n';
    else
      do_swap(a, b);
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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, Q, K;
int P[100005], R[100005], G[100005];
void precalc() {
  for (int i = 1; i <= N; i++)
    if (!G[i]) {
      int b = i;
      for (int j = 0; j < K; j++) b = P[b];
      for (int j = i; !G[j]; j = P[j], b = P[b]) G[j] = b;
    }
}
int get(int n, int step) {
  while (step >= K) {
    n = G[n];
    step -= K;
  }
  while (step--) {
    n = P[n];
  }
  return n;
}
void refresh(int n) {
  int m = n;
  for (int i = 0; i < K; i++) m = P[m];
  G[n] = m;
  for (int i = 0; i < K; i++) {
    n = R[n];
    m = R[m];
    G[n] = m;
  }
}
void do_swap(int a, int b) {
  swap(P[a], P[b]);
  swap(R[P[a]], R[P[b]]);
  refresh(a);
  refresh(b);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> Q;
  K = sqrt(N);
  for (int i = 1; i <= N; i++) {
    cin >> P[i];
    R[P[i]] = i;
  }
  precalc();
  while (Q--) {
    int t, a, b;
    cin >> t >> a >> b;
    if (t == 2)
      cout << get(a, b) << '\n';
    else
      do_swap(a, b);
  }
}

```