#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T re() {
  T N = 0;
  char c = getchar();
  bool neg = 0;
  for (; c < '0' || c > '9'; c = getchar()) neg |= c == '-';
  for (; c >= '0' && c <= '9'; c = getchar()) N = (N << 3) + (N << 1) + c - '0';
  return neg ? -N : N;
}
const int MX = 1e5;
const int SQRTN = 320;
int n, q;
int p[MX + 5];
int rev[MX + 5];
int a[MX + 5];
int main() {
  n = re<int>();
  q = re<int>();
  for (int i = 1; i <= n; i++) rev[p[i] = re<int>()] = i;
  for (int i = 1; i <= n; i++) {
    a[i] = i;
    for (int j = 0; j < SQRTN; j++) a[i] = p[a[i]];
  }
  for (int qq = 0, t, x, y; qq < q; qq++) {
    t = re<int>();
    x = re<int>();
    y = re<int>();
    if (t == 1) {
      swap(rev[p[x]], rev[p[y]]);
      swap(p[x], p[y]);
      for (int ff = 0, u; ff < 2; ff++) {
        u = x;
        for (int i = 0; i < SQRTN; i++) u = p[u];
        for (int i = 0; i < SQRTN; i++) {
          a[x] = u;
          x = rev[x];
          u = rev[u];
        }
        swap(x, y);
      }
    } else {
      for (; y >= SQRTN; y -= SQRTN) x = a[x];
      for (; y > 0; --y) x = p[x];
      printf("%d\n", x);
    }
  }
  return 0;
}

### Prompt

Please formulate a Cpp solution 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>
using namespace std;
template <class T>
inline T re() {
  T N = 0;
  char c = getchar();
  bool neg = 0;
  for (; c < '0' || c > '9'; c = getchar()) neg |= c == '-';
  for (; c >= '0' && c <= '9'; c = getchar()) N = (N << 3) + (N << 1) + c - '0';
  return neg ? -N : N;
}
const int MX = 1e5;
const int SQRTN = 320;
int n, q;
int p[MX + 5];
int rev[MX + 5];
int a[MX + 5];
int main() {
  n = re<int>();
  q = re<int>();
  for (int i = 1; i <= n; i++) rev[p[i] = re<int>()] = i;
  for (int i = 1; i <= n; i++) {
    a[i] = i;
    for (int j = 0; j < SQRTN; j++) a[i] = p[a[i]];
  }
  for (int qq = 0, t, x, y; qq < q; qq++) {
    t = re<int>();
    x = re<int>();
    y = re<int>();
    if (t == 1) {
      swap(rev[p[x]], rev[p[y]]);
      swap(p[x], p[y]);
      for (int ff = 0, u; ff < 2; ff++) {
        u = x;
        for (int i = 0; i < SQRTN; i++) u = p[u];
        for (int i = 0; i < SQRTN; i++) {
          a[x] = u;
          x = rev[x];
          u = rev[u];
        }
        swap(x, y);
      }
    } else {
      for (; y >= SQRTN; y -= SQRTN) x = a[x];
      for (; y > 0; --y) x = p[x];
      printf("%d\n", x);
    }
  }
  return 0;
}

```