#include <bits/stdc++.h>
using namespace std;
template <class T>
void minn(T& a, T b) {
  if (a > b) a = b;
}
template <class T>
void maxx(T& a, T b) {
  if (a < b) a = b;
}
void debug() { return; }
template <class H, class... T>
void debug(H a, T... b) {
  cerr << a;
  debug(b...);
}
void is() { return; }
template <class H, class... T>
void is(H& a, T&... b) {
  cin >> a;
  is(b...);
}
void os() { return; }
template <class H, class... T>
void os(H a, T... b) {
  cout << a;
  os(b...);
}
const int mod = 1e9 + 7;
const int N = 2e5 + 5;
int n, q;
int nx[N], pr[N], nx100[N];
void sw(int x, int y) {
  swap(nx[x], nx[y]);
  pr[nx[x]] = x;
  pr[nx[y]] = y;
  int k = x;
  for (int i = 1; i <= 100; i++) k = pr[k];
  for (int i = 1; i <= 105; i++) {
    nx100[k] = x;
    x = nx[x];
    k = nx[k];
  }
  k = y;
  for (int i = 1; i <= 100; i++) k = pr[k];
  for (int i = 1; i <= 105; i++) {
    nx100[k] = y;
    y = nx[y];
    k = nx[k];
  }
}
void solve() {
  is(n, q);
  for (int i = 1; i <= n; i++) {
    is(nx[i]);
    pr[nx[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    int k = i;
    for (int j = 1; j <= 100; j++) {
      k = nx[k];
    }
    nx100[i] = k;
  }
  while (q--) {
    int t, x, y;
    is(t, x, y);
    if (t == 2) {
      int k = x;
      while (y >= 100) {
        y -= 100;
        k = nx100[k];
      }
      while (y--) k = nx[k];
      os(k, '\n');
    } else {
      sw(x, y);
    }
  }
}
signed main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int T = 1;
  while (T--) {
    solve();
  }
}

### 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;
template <class T>
void minn(T& a, T b) {
  if (a > b) a = b;
}
template <class T>
void maxx(T& a, T b) {
  if (a < b) a = b;
}
void debug() { return; }
template <class H, class... T>
void debug(H a, T... b) {
  cerr << a;
  debug(b...);
}
void is() { return; }
template <class H, class... T>
void is(H& a, T&... b) {
  cin >> a;
  is(b...);
}
void os() { return; }
template <class H, class... T>
void os(H a, T... b) {
  cout << a;
  os(b...);
}
const int mod = 1e9 + 7;
const int N = 2e5 + 5;
int n, q;
int nx[N], pr[N], nx100[N];
void sw(int x, int y) {
  swap(nx[x], nx[y]);
  pr[nx[x]] = x;
  pr[nx[y]] = y;
  int k = x;
  for (int i = 1; i <= 100; i++) k = pr[k];
  for (int i = 1; i <= 105; i++) {
    nx100[k] = x;
    x = nx[x];
    k = nx[k];
  }
  k = y;
  for (int i = 1; i <= 100; i++) k = pr[k];
  for (int i = 1; i <= 105; i++) {
    nx100[k] = y;
    y = nx[y];
    k = nx[k];
  }
}
void solve() {
  is(n, q);
  for (int i = 1; i <= n; i++) {
    is(nx[i]);
    pr[nx[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    int k = i;
    for (int j = 1; j <= 100; j++) {
      k = nx[k];
    }
    nx100[i] = k;
  }
  while (q--) {
    int t, x, y;
    is(t, x, y);
    if (t == 2) {
      int k = x;
      while (y >= 100) {
        y -= 100;
        k = nx100[k];
      }
      while (y--) k = nx[k];
      os(k, '\n');
    } else {
      sw(x, y);
    }
  }
}
signed main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int T = 1;
  while (T--) {
    solve();
  }
}

```