#include <bits/stdc++.h>
using namespace std;
template <typename _T>
void rd(_T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9')
    x = (x << 3) + (x << 1) + (s ^ 48), s = getchar();
  x *= f;
}
template <typename _T>
void write(_T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
  return;
}
const int N = 100005, SQ = sqrt(N);
int p[N], pinv[N], pa[N], nx[N];
int main() {
  int n, q;
  rd(n), rd(q);
  for (register int i(1); i <= n; ++i) rd(p[i]), pinv[p[i]] = i;
  for (register int i(1); i <= n; ++i) {
    int a = i;
    for (register int j(1); j <= SQ; ++j) a = p[a];
    nx[i] = a;
  }
  while (q--) {
    int t, x, y, i, k;
    rd(t);
    if (t == 1) {
      rd(x), rd(y);
      swap(p[x], p[y]);
      pinv[p[x]] = x;
      pinv[p[y]] = y;
      int a = x;
      for (register int j(1); j <= SQ; ++j) a = p[a];
      nx[x] = a;
      for (register int i(1); i <= SQ; ++i) {
        x = pinv[x];
        a = pinv[a];
        nx[x] = a;
      }
      a = y;
      for (register int i(1); i <= SQ; ++i) a = p[a];
      nx[y] = a;
      for (register int i(1); i <= SQ; ++i) {
        y = pinv[y];
        a = pinv[a];
        nx[y] = a;
      }
    } else {
      rd(i), rd(k);
      while (k >= SQ) i = nx[i], k -= SQ;
      while (k--) i = p[i];
      printf("%d\n", i);
    }
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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 <typename _T>
void rd(_T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9')
    x = (x << 3) + (x << 1) + (s ^ 48), s = getchar();
  x *= f;
}
template <typename _T>
void write(_T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
  return;
}
const int N = 100005, SQ = sqrt(N);
int p[N], pinv[N], pa[N], nx[N];
int main() {
  int n, q;
  rd(n), rd(q);
  for (register int i(1); i <= n; ++i) rd(p[i]), pinv[p[i]] = i;
  for (register int i(1); i <= n; ++i) {
    int a = i;
    for (register int j(1); j <= SQ; ++j) a = p[a];
    nx[i] = a;
  }
  while (q--) {
    int t, x, y, i, k;
    rd(t);
    if (t == 1) {
      rd(x), rd(y);
      swap(p[x], p[y]);
      pinv[p[x]] = x;
      pinv[p[y]] = y;
      int a = x;
      for (register int j(1); j <= SQ; ++j) a = p[a];
      nx[x] = a;
      for (register int i(1); i <= SQ; ++i) {
        x = pinv[x];
        a = pinv[a];
        nx[x] = a;
      }
      a = y;
      for (register int i(1); i <= SQ; ++i) a = p[a];
      nx[y] = a;
      for (register int i(1); i <= SQ; ++i) {
        y = pinv[y];
        a = pinv[a];
        nx[y] = a;
      }
    } else {
      rd(i), rd(k);
      while (k >= SQ) i = nx[i], k -= SQ;
      while (k--) i = p[i];
      printf("%d\n", i);
    }
  }
  return 0;
}

```