#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int ans = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = (ans << 1) + (ans << 3) + c - '0';
    c = getchar();
  }
  return ans * f;
}
inline void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x / 10) write(x / 10);
  putchar((char)(x % 10) + '0');
}
template <typename T>
inline T Abs(T a) {
  return a > 0 ? a : -a;
};
template <typename T, typename TT>
inline T Min(T a, TT b) {
  return a < b ? a : b;
}
template <typename T, typename TT>
inline T Max(T a, TT b) {
  return a < b ? b : a;
}
const int N = 1e5 + 5, BLOCK = 144;
int n, q, up1[N], up2[N], dn[N];
inline int climb(int x) {
  for (int i = 1; i <= BLOCK; ++i) x = up1[x];
  return x;
}
signed main() {
  n = read();
  q = read();
  for (int i = 1; i <= n; ++i) {
    up1[i] = read();
    dn[up1[i]] = i;
  }
  for (int i = 1; i <= n; ++i) up2[i] = climb(i);
  while (q--) {
    int opt = read();
    if (opt == 1) {
      int x = read(), y = read();
      swap(dn[up1[x]], dn[up1[y]]);
      swap(up1[x], up1[y]);
      int px = climb(x), py = climb(y);
      for (int i = 0; i <= BLOCK; ++i) {
        up2[x] = px, up2[y] = py;
        x = dn[x], px = dn[px];
        y = dn[y], py = dn[py];
      }
    } else {
      int x = read(), k = read();
      while (k >= BLOCK) x = up2[x], k -= BLOCK;
      while (k) x = up1[x], --k;
      write(x);
      puts("");
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
inline int read() {
  int ans = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = (ans << 1) + (ans << 3) + c - '0';
    c = getchar();
  }
  return ans * f;
}
inline void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x / 10) write(x / 10);
  putchar((char)(x % 10) + '0');
}
template <typename T>
inline T Abs(T a) {
  return a > 0 ? a : -a;
};
template <typename T, typename TT>
inline T Min(T a, TT b) {
  return a < b ? a : b;
}
template <typename T, typename TT>
inline T Max(T a, TT b) {
  return a < b ? b : a;
}
const int N = 1e5 + 5, BLOCK = 144;
int n, q, up1[N], up2[N], dn[N];
inline int climb(int x) {
  for (int i = 1; i <= BLOCK; ++i) x = up1[x];
  return x;
}
signed main() {
  n = read();
  q = read();
  for (int i = 1; i <= n; ++i) {
    up1[i] = read();
    dn[up1[i]] = i;
  }
  for (int i = 1; i <= n; ++i) up2[i] = climb(i);
  while (q--) {
    int opt = read();
    if (opt == 1) {
      int x = read(), y = read();
      swap(dn[up1[x]], dn[up1[y]]);
      swap(up1[x], up1[y]);
      int px = climb(x), py = climb(y);
      for (int i = 0; i <= BLOCK; ++i) {
        up2[x] = px, up2[y] = py;
        x = dn[x], px = dn[px];
        y = dn[y], py = dn[py];
      }
    } else {
      int x = read(), k = read();
      while (k >= BLOCK) x = up2[x], k -= BLOCK;
      while (k) x = up1[x], --k;
      write(x);
      puts("");
    }
  }
  return 0;
}

```