#include <bits/stdc++.h>
using namespace std;
const int jump = 300;
const int MX = 1e5 + 5;
int p[MX], ip[MX], nxt[MX];
void overwrite(int pos, int finish, int turns) {
  if (turns == 0) {
    return;
  }
  nxt[pos] = finish;
  overwrite(ip[pos], ip[finish], turns - 1);
}
void update(int pos) {
  int finish = pos;
  for (int i = 0; i < jump; i++) {
    finish = p[finish];
  }
  overwrite(pos, finish, jump + 5);
}
void runswap(int x, int y) {
  ip[p[x]] = y;
  ip[p[y]] = x;
  swap(p[x], p[y]);
  update(p[x]);
  update(p[y]);
}
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    --p[i];
    ip[p[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    runswap(i, i);
  }
  for (int j = 0; j < q; j++) {
    int v;
    cin >> v;
    if (v == 1) {
      int x, y;
      cin >> x >> y;
      --x;
      --y;
      runswap(x, y);
    } else {
      int i, k;
      cin >> i >> k;
      --i;
      while (k >= jump) {
        k -= jump;
        i = nxt[i];
      }
      while (k > 0) {
        i = p[i];
        k--;
      }
      cout << i + 1 << endl;
    }
  }
  return 0;
}

### 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;
const int jump = 300;
const int MX = 1e5 + 5;
int p[MX], ip[MX], nxt[MX];
void overwrite(int pos, int finish, int turns) {
  if (turns == 0) {
    return;
  }
  nxt[pos] = finish;
  overwrite(ip[pos], ip[finish], turns - 1);
}
void update(int pos) {
  int finish = pos;
  for (int i = 0; i < jump; i++) {
    finish = p[finish];
  }
  overwrite(pos, finish, jump + 5);
}
void runswap(int x, int y) {
  ip[p[x]] = y;
  ip[p[y]] = x;
  swap(p[x], p[y]);
  update(p[x]);
  update(p[y]);
}
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    --p[i];
    ip[p[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    runswap(i, i);
  }
  for (int j = 0; j < q; j++) {
    int v;
    cin >> v;
    if (v == 1) {
      int x, y;
      cin >> x >> y;
      --x;
      --y;
      runswap(x, y);
    } else {
      int i, k;
      cin >> i >> k;
      --i;
      while (k >= jump) {
        k -= jump;
        i = nxt[i];
      }
      while (k > 0) {
        i = p[i];
        k--;
      }
      cout << i + 1 << endl;
    }
  }
  return 0;
}

```