#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = (int)1e5 + 5;
int _ = 1;
int n, q;
int a[N], pos[N];
int nex[N], blo;
int calc(int x) {
  for (int i = 1; i <= blo; i++) x = a[x];
  return x;
}
void work() {
  scanf("%d%d", &n, &q);
  blo = sqrt(n) + 1;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) pos[a[i]] = i;
  for (int i = 1; i <= n; i++) nex[i] = calc(i);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int x, y;
      scanf("%d%d", &x, &y);
      swap(pos[a[x]], pos[a[y]]);
      swap(a[x], a[y]);
      nex[x] = calc(x);
      nex[y] = calc(y);
      for (int i = 1; i <= blo; i++) {
        x = pos[x];
        y = pos[y];
      }
      int ex = calc(x), ey = calc(y);
      for (int i = 1; i <= blo; i++) {
        nex[x] = ex;
        nex[y] = ey;
        x = a[x];
        y = a[y];
        ex = a[ex];
        ey = a[ey];
      }
    } else {
      int x, y;
      scanf("%d%d", &x, &y);
      while (y >= blo) {
        x = nex[x];
        y -= blo;
      }
      while (y) {
        x = a[x];
        y--;
      }
      printf("%d\n", x);
    }
  }
}
int main() {
  while (_--) {
    work();
  }
  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;
using ll = long long;
const int N = (int)1e5 + 5;
int _ = 1;
int n, q;
int a[N], pos[N];
int nex[N], blo;
int calc(int x) {
  for (int i = 1; i <= blo; i++) x = a[x];
  return x;
}
void work() {
  scanf("%d%d", &n, &q);
  blo = sqrt(n) + 1;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) pos[a[i]] = i;
  for (int i = 1; i <= n; i++) nex[i] = calc(i);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int x, y;
      scanf("%d%d", &x, &y);
      swap(pos[a[x]], pos[a[y]]);
      swap(a[x], a[y]);
      nex[x] = calc(x);
      nex[y] = calc(y);
      for (int i = 1; i <= blo; i++) {
        x = pos[x];
        y = pos[y];
      }
      int ex = calc(x), ey = calc(y);
      for (int i = 1; i <= blo; i++) {
        nex[x] = ex;
        nex[y] = ey;
        x = a[x];
        y = a[y];
        ex = a[ex];
        ey = a[ey];
      }
    } else {
      int x, y;
      scanf("%d%d", &x, &y);
      while (y >= blo) {
        x = nex[x];
        y -= blo;
      }
      while (y) {
        x = a[x];
        y--;
      }
      printf("%d\n", x);
    }
  }
}
int main() {
  while (_--) {
    work();
  }
  return 0;
}

```