#include <bits/stdc++.h>
using namespace std;
using cd = complex<double>;
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
const long long N = 2e5 + 5;
long long n, q, a[N], sz = 150, l[N], r[N], R[N];
void up(long long x) {
  long long pre = x, sx = r[x];
  for (long long i = 1; i < sz; i++) pre = l[pre];
  for (long long i = 1; i <= sz; i++) {
    R[pre] = sx;
    sx = r[sx];
    pre = r[pre];
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> r[i], l[r[i]] = i;
  for (long long i = 1; i <= n; i++) up(i);
  while (q--) {
    long long op;
    cin >> op;
    if (op == 1) {
      long long x, y;
      cin >> x >> y;
      swap(r[x], r[y]);
      l[r[x]] = x;
      l[r[y]] = y;
      up(x), up(y);
    } else {
      long long x, k;
      cin >> x >> k;
      while (k >= sz) k -= sz, x = R[x];
      while (k) k--, x = r[x];
      cout << x << endl;
    }
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
using cd = complex<double>;
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
const long long N = 2e5 + 5;
long long n, q, a[N], sz = 150, l[N], r[N], R[N];
void up(long long x) {
  long long pre = x, sx = r[x];
  for (long long i = 1; i < sz; i++) pre = l[pre];
  for (long long i = 1; i <= sz; i++) {
    R[pre] = sx;
    sx = r[sx];
    pre = r[pre];
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> r[i], l[r[i]] = i;
  for (long long i = 1; i <= n; i++) up(i);
  while (q--) {
    long long op;
    cin >> op;
    if (op == 1) {
      long long x, y;
      cin >> x >> y;
      swap(r[x], r[y]);
      l[r[x]] = x;
      l[r[y]] = y;
      up(x), up(y);
    } else {
      long long x, k;
      cin >> x >> k;
      while (k >= sz) k -= sz, x = R[x];
      while (k) k--, x = r[x];
      cout << x << endl;
    }
  }
}

```