#include <bits/stdc++.h>
using namespace std;
int way[100005][9];
int mem[(1 << (9 + 2))];
int p[100005];
int rev[100005];
void upd(int x) {
  int grv = (1 << (9 + 1));
  int tg = x;
  for (int i = 0; i <= (1 << (9 - 1)); i++) {
    mem[grv + i] = tg;
    tg = p[tg];
  }
  tg = x;
  for (int i = 0; i <= (1 << (9 - 1)); i++) {
    for (int j = 0; j < 9; j++) {
      way[tg][j] = mem[grv + (1 << j)];
    }
    tg = rev[tg];
    grv--;
    mem[grv] = tg;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    rev[p[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    int tg = i, np = 0;
    for (int j = 1; j <= (1 << (9 - 1)); j++) {
      tg = p[tg];
      if (j == (1 << np)) {
        way[i][np] = tg;
        np++;
      }
    }
  }
  while (q > 0) {
    q--;
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      x--;
      y--;
      swap(p[x], p[y]);
      rev[p[x]] = x;
      rev[p[y]] = y;
      upd(x);
      upd(y);
    } else {
      x--;
      int res = x;
      int cp = 9 - 1;
      while (y > 0) {
        while (y < (1 << cp)) {
          cp--;
        }
        y -= (1 << cp);
        res = way[res][cp];
      }
      cout << res + 1 << '\n';
    }
  }
  return 0;
}

### Prompt

Generate 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;
int way[100005][9];
int mem[(1 << (9 + 2))];
int p[100005];
int rev[100005];
void upd(int x) {
  int grv = (1 << (9 + 1));
  int tg = x;
  for (int i = 0; i <= (1 << (9 - 1)); i++) {
    mem[grv + i] = tg;
    tg = p[tg];
  }
  tg = x;
  for (int i = 0; i <= (1 << (9 - 1)); i++) {
    for (int j = 0; j < 9; j++) {
      way[tg][j] = mem[grv + (1 << j)];
    }
    tg = rev[tg];
    grv--;
    mem[grv] = tg;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    rev[p[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    int tg = i, np = 0;
    for (int j = 1; j <= (1 << (9 - 1)); j++) {
      tg = p[tg];
      if (j == (1 << np)) {
        way[i][np] = tg;
        np++;
      }
    }
  }
  while (q > 0) {
    q--;
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      x--;
      y--;
      swap(p[x], p[y]);
      rev[p[x]] = x;
      rev[p[y]] = y;
      upd(x);
      upd(y);
    } else {
      x--;
      int res = x;
      int cp = 9 - 1;
      while (y > 0) {
        while (y < (1 << cp)) {
          cp--;
        }
        y -= (1 << cp);
        res = way[res][cp];
      }
      cout << res + 1 << '\n';
    }
  }
  return 0;
}

```