#include <bits/stdc++.h>
using namespace std;
string _reset = "\u001b[0m", _yellow = "\u001b[33m", _bold = "\u001b[1m";
void DBG() { cerr << "]" << _reset << endl; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << to_string(h);
  if (sizeof...(t)) cerr << ", ";
  DBG(t...);
}
const int INF = 1e9 + 7, N = 1e5 + 7, K = 320;
int p[N], jump[N], ip[N];
bool vis[N];
signed main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  ;
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    ip[p[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    jump[i] = i;
    for (int j = 0; j < K; j++) jump[i] = p[jump[i]];
  }
  auto fix = [&](vector<int>& v) {
    assert(v.size());
    jump[v[0]] = v[0];
    for (int j = 0; j < K; j++) jump[v[0]] = p[jump[v[0]]];
    for (int i = 1; i < int(v.size()); i++) jump[v[i]] = p[jump[v[i - 1]]];
  };
  while (q--) {
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      x--, y--;
      vector<int> fix1, fix2;
      {
        int u = x;
        fix1.push_back(u);
        while (ip[u] != x && ip[u] != y && int(fix1.size()) < K) {
          u = ip[u];
          fix1.push_back(u);
        }
        reverse((fix1).begin(), (fix1).end());
      }
      {
        int u = y;
        fix2.push_back(u);
        while (ip[u] != x && ip[u] != y && int(fix2.size()) < K) {
          u = ip[u];
          fix2.push_back(u);
        }
        reverse((fix2).begin(), (fix2).end());
      }
      swap(p[x], p[y]);
      swap(ip[p[x]], ip[p[y]]);
      fix(fix1), fix(fix2);
    } else {
      x--;
      while (y >= K) {
        x = jump[x];
        y -= K;
      }
      while (y) {
        x = p[x];
        y--;
      }
      cout << x + 1 << '\n';
    }
  }
}

### 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;
string _reset = "\u001b[0m", _yellow = "\u001b[33m", _bold = "\u001b[1m";
void DBG() { cerr << "]" << _reset << endl; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << to_string(h);
  if (sizeof...(t)) cerr << ", ";
  DBG(t...);
}
const int INF = 1e9 + 7, N = 1e5 + 7, K = 320;
int p[N], jump[N], ip[N];
bool vis[N];
signed main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  ;
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    ip[p[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    jump[i] = i;
    for (int j = 0; j < K; j++) jump[i] = p[jump[i]];
  }
  auto fix = [&](vector<int>& v) {
    assert(v.size());
    jump[v[0]] = v[0];
    for (int j = 0; j < K; j++) jump[v[0]] = p[jump[v[0]]];
    for (int i = 1; i < int(v.size()); i++) jump[v[i]] = p[jump[v[i - 1]]];
  };
  while (q--) {
    int t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      x--, y--;
      vector<int> fix1, fix2;
      {
        int u = x;
        fix1.push_back(u);
        while (ip[u] != x && ip[u] != y && int(fix1.size()) < K) {
          u = ip[u];
          fix1.push_back(u);
        }
        reverse((fix1).begin(), (fix1).end());
      }
      {
        int u = y;
        fix2.push_back(u);
        while (ip[u] != x && ip[u] != y && int(fix2.size()) < K) {
          u = ip[u];
          fix2.push_back(u);
        }
        reverse((fix2).begin(), (fix2).end());
      }
      swap(p[x], p[y]);
      swap(ip[p[x]], ip[p[y]]);
      fix(fix1), fix(fix2);
    } else {
      x--;
      while (y >= K) {
        x = jump[x];
        y -= K;
      }
      while (y) {
        x = p[x];
        y--;
      }
      cout << x + 1 << '\n';
    }
  }
}

```