#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
int i, j, k, n, m, t, x, y, z;
int l[100500], r[100500], l300[100500], r300[100500], k2;
vector<pair<int, int> > v1, v2;
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> t;
  for (i = 1; i <= n; i++) {
    cin >> r[i];
    l[r[i]] = i;
  }
  for (i = 1; i <= n; i++) {
    k = i;
    for (j = 1; j <= 150; j++) {
      k = r[k];
    }
    r300[i] = k;
    l300[k] = i;
  }
  while (t--) {
    cin >> x >> y >> z;
    if (x == 1) {
      swap(r[y], r[z]);
      l[r[y]] = y;
      l[r[z]] = z;
      k = y;
      for (i = 1; i < 150; i++) {
        k = l[k];
      }
      v1.clear();
      k2 = r[y];
      for (i = 1; i <= 150; i++) {
        v1.push_back({k, k2});
        k = r[k];
        k2 = r[k2];
      }
      k = z;
      for (i = 1; i < 150; i++) {
        k = l[k];
      }
      v2.clear();
      k2 = r[z];
      for (i = 1; i <= 150; i++) {
        v2.push_back({k, k2});
        k = r[k];
        k2 = r[k2];
      }
      for (auto [i, j] : v1) {
        r300[i] = j;
        l300[j] = i;
      }
      for (auto [i, j] : v2) {
        r300[i] = j;
        l300[j] = i;
      }
    } else {
      while (z > 150) {
        y = r300[y];
        z -= 150;
      }
      while (z) {
        y = r[y];
        z--;
      }
      cout << y << '\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;
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
int i, j, k, n, m, t, x, y, z;
int l[100500], r[100500], l300[100500], r300[100500], k2;
vector<pair<int, int> > v1, v2;
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> t;
  for (i = 1; i <= n; i++) {
    cin >> r[i];
    l[r[i]] = i;
  }
  for (i = 1; i <= n; i++) {
    k = i;
    for (j = 1; j <= 150; j++) {
      k = r[k];
    }
    r300[i] = k;
    l300[k] = i;
  }
  while (t--) {
    cin >> x >> y >> z;
    if (x == 1) {
      swap(r[y], r[z]);
      l[r[y]] = y;
      l[r[z]] = z;
      k = y;
      for (i = 1; i < 150; i++) {
        k = l[k];
      }
      v1.clear();
      k2 = r[y];
      for (i = 1; i <= 150; i++) {
        v1.push_back({k, k2});
        k = r[k];
        k2 = r[k2];
      }
      k = z;
      for (i = 1; i < 150; i++) {
        k = l[k];
      }
      v2.clear();
      k2 = r[z];
      for (i = 1; i <= 150; i++) {
        v2.push_back({k, k2});
        k = r[k];
        k2 = r[k2];
      }
      for (auto [i, j] : v1) {
        r300[i] = j;
        l300[j] = i;
      }
      for (auto [i, j] : v2) {
        r300[i] = j;
        l300[j] = i;
      }
    } else {
      while (z > 150) {
        y = r300[y];
        z -= 150;
      }
      while (z) {
        y = r[y];
        z--;
      }
      cout << y << '\n';
    }
  }
}

```