#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("popcnt")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const int maxn = 1e5 + 11;
const int BLOCK_SIZE = 500;
int n, q, rev[maxn], p[maxn];
int nxt[maxn];
int compute(int i) {
  int cur = i;
  for (int j = 1; j <= BLOCK_SIZE; j++) {
    cur = p[cur];
  }
  return (nxt[i] = cur);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
    rev[p[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    compute(i);
  }
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int x, y;
      cin >> x >> y;
      swap(rev[p[x]], rev[p[y]]);
      swap(p[x], p[y]);
      int sx = compute(x), sy = compute(y);
      for (int i = 1; i <= BLOCK_SIZE; i++) {
        nxt[x] = sx;
        nxt[y] = sy;
        x = rev[x];
        y = rev[y];
        sx = rev[sx];
        sy = rev[sy];
      }
    } else {
      int i, k;
      cin >> i >> k;
      int tmp = nxt[i];
      while (k > 0 && i != tmp) {
        i = p[i];
        k--;
      }
      while (k >= BLOCK_SIZE) {
        i = nxt[i];
        k -= BLOCK_SIZE;
      }
      while (k--) {
        i = p[i];
      }
      cout << i << '\n';
    }
  }
  return 0;
}

### Prompt

Your task is to create 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("popcnt")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const int maxn = 1e5 + 11;
const int BLOCK_SIZE = 500;
int n, q, rev[maxn], p[maxn];
int nxt[maxn];
int compute(int i) {
  int cur = i;
  for (int j = 1; j <= BLOCK_SIZE; j++) {
    cur = p[cur];
  }
  return (nxt[i] = cur);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
    rev[p[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    compute(i);
  }
  while (q--) {
    int type;
    cin >> type;
    if (type == 1) {
      int x, y;
      cin >> x >> y;
      swap(rev[p[x]], rev[p[y]]);
      swap(p[x], p[y]);
      int sx = compute(x), sy = compute(y);
      for (int i = 1; i <= BLOCK_SIZE; i++) {
        nxt[x] = sx;
        nxt[y] = sy;
        x = rev[x];
        y = rev[y];
        sx = rev[sx];
        sy = rev[sy];
      }
    } else {
      int i, k;
      cin >> i >> k;
      int tmp = nxt[i];
      while (k > 0 && i != tmp) {
        i = p[i];
        k--;
      }
      while (k >= BLOCK_SIZE) {
        i = nxt[i];
        k -= BLOCK_SIZE;
      }
      while (k--) {
        i = p[i];
      }
      cout << i << '\n';
    }
  }
  return 0;
}

```