#include <bits/stdc++.h>
using namespace std;
using LL = long long;
using ULL = unsigned long long;
using ld = long double;
const long long N = 2e5 + 10, mod = 1e9 + 7;
const LL INF = 1e16;
long long dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
long long ne[N], pre[N], p[N];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n, q;
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    cin >> p[i];
    pre[p[i]] = i;
  }
  long long len = sqrt(n);
  for (long long i = 1; i <= n; i++) {
    long long x = i;
    for (long long j = 1; j <= len; j++) {
      x = p[x];
    }
    ne[i] = x;
  }
  while (q--) {
    long long op, x, y;
    cin >> op >> x >> y;
    if (op == 1) {
      swap(p[x], p[y]);
      pre[p[x]] = x;
      pre[p[y]] = y;
      long long a = x;
      for (long long j = 1; j <= len; j++) a = p[a];
      ne[x] = a;
      for (long long j = 1; j <= len; j++) {
        x = pre[x];
        a = pre[a];
        ne[x] = a;
      }
      a = y;
      for (long long j = 1; j <= len; j++) a = p[a];
      ne[y] = a;
      for (long long j = 1; j <= len; j++) {
        y = pre[y];
        a = pre[a];
        ne[y] = a;
      }
    } else {
      while (y >= len) {
        x = ne[x];
        y -= len;
      }
      for (long long i = 1; i <= y; i++) {
        x = p[x];
      }
      cout << x << '\n';
    }
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
using ULL = unsigned long long;
using ld = long double;
const long long N = 2e5 + 10, mod = 1e9 + 7;
const LL INF = 1e16;
long long dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
long long ne[N], pre[N], p[N];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n, q;
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) {
    cin >> p[i];
    pre[p[i]] = i;
  }
  long long len = sqrt(n);
  for (long long i = 1; i <= n; i++) {
    long long x = i;
    for (long long j = 1; j <= len; j++) {
      x = p[x];
    }
    ne[i] = x;
  }
  while (q--) {
    long long op, x, y;
    cin >> op >> x >> y;
    if (op == 1) {
      swap(p[x], p[y]);
      pre[p[x]] = x;
      pre[p[y]] = y;
      long long a = x;
      for (long long j = 1; j <= len; j++) a = p[a];
      ne[x] = a;
      for (long long j = 1; j <= len; j++) {
        x = pre[x];
        a = pre[a];
        ne[x] = a;
      }
      a = y;
      for (long long j = 1; j <= len; j++) a = p[a];
      ne[y] = a;
      for (long long j = 1; j <= len; j++) {
        y = pre[y];
        a = pre[a];
        ne[y] = a;
      }
    } else {
      while (y >= len) {
        x = ne[x];
        y -= len;
      }
      for (long long i = 1; i <= y; i++) {
        x = p[x];
      }
      cout << x << '\n';
    }
  }
}

```