#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 1e9 + 7;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
int n, q;
int l[N], r[N], ans_sqrt[N];
int Q = 100;
void solve() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &r[i]);
    l[r[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    int p = i;
    for (int j = 0; j < Q; j++) {
      p = r[p];
    }
    ans_sqrt[i] = p;
  }
  while (q--) {
    int t, x, y;
    scanf("%d%d%d", &t, &x, &y);
    if (t == 1) {
      swap(r[x], r[y]);
      swap(l[r[x]], l[r[y]]);
      vector<int> vv;
      int pp = x;
      for (int p = 0; p < Q; p++) {
        pp = r[pp];
        vv.push_back(pp);
      }
      while (!vv.empty()) {
        ans_sqrt[x] = vv.back();
        vv.pop_back();
        x = l[x];
      }
      pp = y;
      for (int p = 0; p < Q; p++) {
        pp = r[pp];
        vv.push_back(pp);
      }
      while (!vv.empty()) {
        ans_sqrt[y] = vv.back();
        vv.pop_back();
        y = l[y];
      }
    } else {
      int ans = x;
      while (y >= Q) {
        y -= Q;
        ans = ans_sqrt[ans];
      }
      while (y--) {
        ans = r[ans];
      }
      printf("%d\n", ans);
    }
  }
}
int main() {
  int T = 1;
  while (T--) {
    solve();
  }
  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>
using namespace std;
const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 1e9 + 7;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
int n, q;
int l[N], r[N], ans_sqrt[N];
int Q = 100;
void solve() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &r[i]);
    l[r[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    int p = i;
    for (int j = 0; j < Q; j++) {
      p = r[p];
    }
    ans_sqrt[i] = p;
  }
  while (q--) {
    int t, x, y;
    scanf("%d%d%d", &t, &x, &y);
    if (t == 1) {
      swap(r[x], r[y]);
      swap(l[r[x]], l[r[y]]);
      vector<int> vv;
      int pp = x;
      for (int p = 0; p < Q; p++) {
        pp = r[pp];
        vv.push_back(pp);
      }
      while (!vv.empty()) {
        ans_sqrt[x] = vv.back();
        vv.pop_back();
        x = l[x];
      }
      pp = y;
      for (int p = 0; p < Q; p++) {
        pp = r[pp];
        vv.push_back(pp);
      }
      while (!vv.empty()) {
        ans_sqrt[y] = vv.back();
        vv.pop_back();
        y = l[y];
      }
    } else {
      int ans = x;
      while (y >= Q) {
        y -= Q;
        ans = ans_sqrt[ans];
      }
      while (y--) {
        ans = r[ans];
      }
      printf("%d\n", ans);
    }
  }
}
int main() {
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```