#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, q;
int pre[N], nxt[N];
int sq;
int num;
int cnt;
int now;
int c[N], f[N], t[N];
vector<int> need_resolve;
int calc_nxt(int x) {
  int k = sq;
  for (; k >= 4; k -= 4) {
    x = nxt[nxt[nxt[nxt[x]]]];
  }
  for (; k--;) {
    x = nxt[x];
  }
  return x;
}
int calc_pre(int x) {
  int k = sq;
  for (; k >= 4; k -= 4) {
    x = pre[pre[pre[pre[x]]]];
  }
  for (; k--;) {
    x = pre[x];
  }
  return x;
}
int main() {
  scanf("%d%d", &n, &q);
  sq = sqrt(n);
  sq = min(sq, 100);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", nxt + i);
    pre[nxt[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    f[i] = calc_nxt(i);
  }
  while (q--) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    int nx = nxt[x], ny = nxt[y];
    if (op == 1) {
      need_resolve.clear();
      need_resolve.emplace_back(calc_pre(x));
      need_resolve.emplace_back(calc_pre(y));
      need_resolve.emplace_back(calc_pre(nx));
      need_resolve.emplace_back(calc_pre(ny));
      nxt[x] = ny;
      pre[ny] = x;
      nxt[y] = nx;
      pre[nx] = y;
      need_resolve.emplace_back(calc_pre(nx));
      need_resolve.emplace_back(calc_pre(ny));
      for (auto u : need_resolve) {
        t[1] = u;
        for (int i = 2; i <= 2 * sq + 2; ++i) {
          t[i] = nxt[t[i - 1]];
        }
        for (int i = 1; i <= sq + 2; ++i) {
          f[t[i]] = t[i + sq];
        }
      }
    } else {
      int ret = x;
      for (; y > sq; y -= sq) {
        ret = f[ret];
      }
      for (; y; --y) {
        ret = nxt[ret];
      }
      printf("%d\n", ret);
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 = 1e5 + 10;
int n, q;
int pre[N], nxt[N];
int sq;
int num;
int cnt;
int now;
int c[N], f[N], t[N];
vector<int> need_resolve;
int calc_nxt(int x) {
  int k = sq;
  for (; k >= 4; k -= 4) {
    x = nxt[nxt[nxt[nxt[x]]]];
  }
  for (; k--;) {
    x = nxt[x];
  }
  return x;
}
int calc_pre(int x) {
  int k = sq;
  for (; k >= 4; k -= 4) {
    x = pre[pre[pre[pre[x]]]];
  }
  for (; k--;) {
    x = pre[x];
  }
  return x;
}
int main() {
  scanf("%d%d", &n, &q);
  sq = sqrt(n);
  sq = min(sq, 100);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", nxt + i);
    pre[nxt[i]] = i;
  }
  for (int i = 1; i <= n; ++i) {
    f[i] = calc_nxt(i);
  }
  while (q--) {
    int op, x, y;
    scanf("%d%d%d", &op, &x, &y);
    int nx = nxt[x], ny = nxt[y];
    if (op == 1) {
      need_resolve.clear();
      need_resolve.emplace_back(calc_pre(x));
      need_resolve.emplace_back(calc_pre(y));
      need_resolve.emplace_back(calc_pre(nx));
      need_resolve.emplace_back(calc_pre(ny));
      nxt[x] = ny;
      pre[ny] = x;
      nxt[y] = nx;
      pre[nx] = y;
      need_resolve.emplace_back(calc_pre(nx));
      need_resolve.emplace_back(calc_pre(ny));
      for (auto u : need_resolve) {
        t[1] = u;
        for (int i = 2; i <= 2 * sq + 2; ++i) {
          t[i] = nxt[t[i - 1]];
        }
        for (int i = 1; i <= sq + 2; ++i) {
          f[t[i]] = t[i + sq];
        }
      }
    } else {
      int ret = x;
      for (; y > sq; y -= sq) {
        ret = f[ret];
      }
      for (; y; --y) {
        ret = nxt[ret];
      }
      printf("%d\n", ret);
    }
  }
  return 0;
}

```