#include <bits/stdc++.h>
template <class T>
inline T read() {
  T r = 0, f = 0;
  char c;
  while (!isdigit(c = getchar())) f |= (c == '-');
  while (isdigit(c)) r = (r << 1) + (r << 3) + (c ^ 48), c = getchar();
  return f ? -r : r;
}
template <class T>
inline void swap(T &a, T &b) {
  T c = a;
  a = b;
  b = c;
}
int n, m, B, p[101101], r[101101], jump[101101];
inline void Jump(int x) {
  int X = x;
  for (int i = 1; i <= B; i++) X = p[X];
  for (int i = 1; i <= B; i++) jump[x] = X, x = r[x], X = r[X];
}
inline void Modify() {
  int x = read<int>();
  int y = read<int>();
  swap(r[p[x]], r[p[y]]);
  swap(p[x], p[y]);
  Jump(x), Jump(y);
}
inline void work() {
  int opt = read<int>();
  if (opt == 1) return Modify();
  int x = read<int>(), k = read<int>();
  while (k >= B) k -= B, x = jump[x];
  while (k) x = p[x], k--;
  printf("%d\n", x);
}
int main() {
  n = read<int>();
  m = read<int>();
  B = sqrt(n);
  for (int i = 1; i <= n; i++) p[i] = read<int>(), r[p[i]] = i;
  for (int i = 1; i <= n; i++) {
    jump[i] = i;
    for (int j = 1; j <= B; j++) jump[i] = p[jump[i]];
  }
  for (int t = 1; t <= m; t++) work();
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
template <class T>
inline T read() {
  T r = 0, f = 0;
  char c;
  while (!isdigit(c = getchar())) f |= (c == '-');
  while (isdigit(c)) r = (r << 1) + (r << 3) + (c ^ 48), c = getchar();
  return f ? -r : r;
}
template <class T>
inline void swap(T &a, T &b) {
  T c = a;
  a = b;
  b = c;
}
int n, m, B, p[101101], r[101101], jump[101101];
inline void Jump(int x) {
  int X = x;
  for (int i = 1; i <= B; i++) X = p[X];
  for (int i = 1; i <= B; i++) jump[x] = X, x = r[x], X = r[X];
}
inline void Modify() {
  int x = read<int>();
  int y = read<int>();
  swap(r[p[x]], r[p[y]]);
  swap(p[x], p[y]);
  Jump(x), Jump(y);
}
inline void work() {
  int opt = read<int>();
  if (opt == 1) return Modify();
  int x = read<int>(), k = read<int>();
  while (k >= B) k -= B, x = jump[x];
  while (k) x = p[x], k--;
  printf("%d\n", x);
}
int main() {
  n = read<int>();
  m = read<int>();
  B = sqrt(n);
  for (int i = 1; i <= n; i++) p[i] = read<int>(), r[p[i]] = i;
  for (int i = 1; i <= n; i++) {
    jump[i] = i;
    for (int j = 1; j <= B; j++) jump[i] = p[jump[i]];
  }
  for (int t = 1; t <= m; t++) work();
  return 0;
}

```