#include <bits/stdc++.h>
const int maxn = 1e5 + 10;
int n, q, m, p[maxn], r[maxn], tar[maxn];
inline int jump(int x, int k) {
  for (int i = 0; i < k; ++i) x = p[x];
  return x;
}
inline int query(int x, int k) {
  while (k >= m) {
    x = tar[x];
    k -= m;
  }
  for (int i = 0; i < k; ++i) x = p[x];
  return x;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", p + i);
    r[p[i]] = i;
  }
  m = ceil(sqrt(n));
  for (int i = 1; i <= n; ++i) tar[i] = jump(i, m);
  int opt, x, y;
  while (q--) {
    scanf("%d%d%d", &opt, &x, &y);
    if (opt == 1) {
      std::swap(p[x], p[y]);
      std::swap(r[p[x]], r[p[y]]);
      tar[x] = jump(x, m);
      for (int i = 0; i < m; ++i) {
        tar[r[x]] = r[tar[x]];
        x = r[x];
      }
      tar[y] = jump(y, m);
      for (int i = 0; i < m; ++i) {
        tar[r[y]] = r[tar[y]];
        y = r[y];
      }
    } else
      printf("%d\n", query(x, y));
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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>
const int maxn = 1e5 + 10;
int n, q, m, p[maxn], r[maxn], tar[maxn];
inline int jump(int x, int k) {
  for (int i = 0; i < k; ++i) x = p[x];
  return x;
}
inline int query(int x, int k) {
  while (k >= m) {
    x = tar[x];
    k -= m;
  }
  for (int i = 0; i < k; ++i) x = p[x];
  return x;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", p + i);
    r[p[i]] = i;
  }
  m = ceil(sqrt(n));
  for (int i = 1; i <= n; ++i) tar[i] = jump(i, m);
  int opt, x, y;
  while (q--) {
    scanf("%d%d%d", &opt, &x, &y);
    if (opt == 1) {
      std::swap(p[x], p[y]);
      std::swap(r[p[x]], r[p[y]]);
      tar[x] = jump(x, m);
      for (int i = 0; i < m; ++i) {
        tar[r[x]] = r[tar[x]];
        x = r[x];
      }
      tar[y] = jump(y, m);
      for (int i = 0; i < m; ++i) {
        tar[r[y]] = r[tar[y]];
        y = r[y];
      }
    } else
      printf("%d\n", query(x, y));
  }
  return 0;
}

```