#include <bits/stdc++.h>
using namespace std;
template <class T>
void print(vector<T> a) {
  int n = a.size();
  for (long long i = 0; i < n; ++i) {
    cout << a[i] << (i == n - 1 ? "\n" : " ");
  }
}
int sum_vector(vector<int> v) { return accumulate(v.begin(), v.end(), 0); }
void sort_vector(vector<int> &v) { sort(v.begin(), v.end()); }
void sort_comp(vector<int> &v, bool func(int, int)) {
  sort(v.begin(), v.end(), func);
}
bool comp(int a, int b) { return a < b; }
struct CustomCompare {
  bool operator()(pair<int, int> lhs, pair<int, int> rhs) {
    return lhs.first > rhs.first;
  }
};
long long gcd(long long a, long long b) {
  a = abs(a);
  b = abs(b);
  while (a) {
    long long temp = a;
    a = b % a;
    b = temp;
  }
  return abs(b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
string binary(long long num) {
  string ans = "";
  do {
    ans = to_string(num % 2) + ans;
    num /= 2;
  } while (num);
  return ans;
}
const int mxn = 5e5 + 7;
const int d = 18;
const int mill = 1e6 + 3;
const long long mod = 998244353;
long long pwr(long long num, long long p) {
  long long res = 1;
  while (p > 0) {
    if (p & 1) res = (res * num) % mod;
    num = (num * num) % mod;
    p /= 2;
  }
  return res;
}
long long inverse(long long num) { return pwr(num, mod - 2); }
long long get_rand(long long n) { return ((rand() << 15) + rand()) % n; }
void solve() {
  int n, q;
  const int step = 400;
  cin >> n >> q;
  vector<int> p(n + 1);
  vector<int> wtf(n + 1);
  vector<int> pws(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    wtf[p[i]] = i;
  }
  int temp;
  for (int i = 1; i <= n; i++) {
    temp = i;
    for (int s = 0; s < step; s++) {
      temp = p[temp];
    }
    pws[i] = temp;
  }
  int type;
  int x, y;
  int idx, k;
  vector<int> helpx(step + 1);
  vector<int> helpy(step + 1);
  for (int qq = 0; qq < q; qq++) {
    cin >> type;
    if (type == 1) {
      cin >> x >> y;
      temp = p[x];
      p[x] = p[y];
      p[y] = temp;
      wtf[p[x]] = x;
      wtf[p[y]] = y;
      helpx[0] = x;
      helpy[0] = y;
      for (int s = 1; s <= step; s++) {
        helpx[s] = p[helpx[s - 1]];
        helpy[s] = p[helpy[s - 1]];
      }
      pws[x] = helpx[step];
      pws[y] = helpy[step];
      for (int s = 1; s <= step; s++) {
        x = wtf[x];
        y = wtf[y];
        pws[x] = helpx[step - s];
        pws[y] = helpy[step - s];
      }
    } else {
      cin >> idx >> k;
      while (k >= step) {
        idx = pws[idx];
        k -= step;
      }
      for (int s = 0; s < k; s++) {
        idx = p[idx];
      }
      cout << idx << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int t = 1;
  for (int ii = 1; ii <= t; ii++) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate 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 print(vector<T> a) {
  int n = a.size();
  for (long long i = 0; i < n; ++i) {
    cout << a[i] << (i == n - 1 ? "\n" : " ");
  }
}
int sum_vector(vector<int> v) { return accumulate(v.begin(), v.end(), 0); }
void sort_vector(vector<int> &v) { sort(v.begin(), v.end()); }
void sort_comp(vector<int> &v, bool func(int, int)) {
  sort(v.begin(), v.end(), func);
}
bool comp(int a, int b) { return a < b; }
struct CustomCompare {
  bool operator()(pair<int, int> lhs, pair<int, int> rhs) {
    return lhs.first > rhs.first;
  }
};
long long gcd(long long a, long long b) {
  a = abs(a);
  b = abs(b);
  while (a) {
    long long temp = a;
    a = b % a;
    b = temp;
  }
  return abs(b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
string binary(long long num) {
  string ans = "";
  do {
    ans = to_string(num % 2) + ans;
    num /= 2;
  } while (num);
  return ans;
}
const int mxn = 5e5 + 7;
const int d = 18;
const int mill = 1e6 + 3;
const long long mod = 998244353;
long long pwr(long long num, long long p) {
  long long res = 1;
  while (p > 0) {
    if (p & 1) res = (res * num) % mod;
    num = (num * num) % mod;
    p /= 2;
  }
  return res;
}
long long inverse(long long num) { return pwr(num, mod - 2); }
long long get_rand(long long n) { return ((rand() << 15) + rand()) % n; }
void solve() {
  int n, q;
  const int step = 400;
  cin >> n >> q;
  vector<int> p(n + 1);
  vector<int> wtf(n + 1);
  vector<int> pws(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    wtf[p[i]] = i;
  }
  int temp;
  for (int i = 1; i <= n; i++) {
    temp = i;
    for (int s = 0; s < step; s++) {
      temp = p[temp];
    }
    pws[i] = temp;
  }
  int type;
  int x, y;
  int idx, k;
  vector<int> helpx(step + 1);
  vector<int> helpy(step + 1);
  for (int qq = 0; qq < q; qq++) {
    cin >> type;
    if (type == 1) {
      cin >> x >> y;
      temp = p[x];
      p[x] = p[y];
      p[y] = temp;
      wtf[p[x]] = x;
      wtf[p[y]] = y;
      helpx[0] = x;
      helpy[0] = y;
      for (int s = 1; s <= step; s++) {
        helpx[s] = p[helpx[s - 1]];
        helpy[s] = p[helpy[s - 1]];
      }
      pws[x] = helpx[step];
      pws[y] = helpy[step];
      for (int s = 1; s <= step; s++) {
        x = wtf[x];
        y = wtf[y];
        pws[x] = helpx[step - s];
        pws[y] = helpy[step - s];
      }
    } else {
      cin >> idx >> k;
      while (k >= step) {
        idx = pws[idx];
        k -= step;
      }
      for (int s = 0; s < k; s++) {
        idx = p[idx];
      }
      cout << idx << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int t = 1;
  for (int ii = 1; ii <= t; ii++) {
    solve();
  }
  return 0;
}

```