#include <bits/stdc++.h>
using namespace std;
const long long inf = 4e18;
const long long N = 2e5 + 10;
template <typename T>
bool ckmin(T& a, T b) {
  return b < a && (a = b, true);
}
template <typename T>
bool ckmax(T& a, T b) {
  return b > a && (a = b, true);
}
void testCase() {
  long long n, m;
  cin >> n >> m;
  vector<long long> p(n), inv_p(n), reach(n);
  long long S = sqrt(0.45 * n) + 0.5;
  for (long long& i : p) cin >> i, i--;
  auto compute = [&](long long x) {
    long long i = x;
    for (long long r = 0; r < S; r++) {
      x = p[x];
    }
    return reach[i] = x;
  };
  for (long long i = 0; i < n; i++) {
    compute(i);
    inv_p[p[i]] = i;
  }
  for (long long i = 0; i < m; i++) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long x, y;
      cin >> x >> y;
      x--, y--;
      swap(inv_p[p[x]], inv_p[p[y]]);
      swap(p[x], p[y]);
      long long sx = compute(x), sy = compute(y);
      for (long long r = 0; r < S; r++) {
        reach[x] = sx, reach[y] = sy;
        x = inv_p[x], y = inv_p[y];
        sx = inv_p[sx], sy = inv_p[sy];
      }
    } else {
      long long x, k;
      cin >> x >> k;
      x--;
      while (k >= S) {
        x = reach[x];
        k -= S;
      }
      while (k--) {
        x = p[x];
      }
      cout << x + 1 << '\n';
    }
  }
}
int32_t main() {
  cin.tie(0)->sync_with_stdio(0);
  long long t_c = 1;
  while (t_c--) testCase();
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 long long inf = 4e18;
const long long N = 2e5 + 10;
template <typename T>
bool ckmin(T& a, T b) {
  return b < a && (a = b, true);
}
template <typename T>
bool ckmax(T& a, T b) {
  return b > a && (a = b, true);
}
void testCase() {
  long long n, m;
  cin >> n >> m;
  vector<long long> p(n), inv_p(n), reach(n);
  long long S = sqrt(0.45 * n) + 0.5;
  for (long long& i : p) cin >> i, i--;
  auto compute = [&](long long x) {
    long long i = x;
    for (long long r = 0; r < S; r++) {
      x = p[x];
    }
    return reach[i] = x;
  };
  for (long long i = 0; i < n; i++) {
    compute(i);
    inv_p[p[i]] = i;
  }
  for (long long i = 0; i < m; i++) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long x, y;
      cin >> x >> y;
      x--, y--;
      swap(inv_p[p[x]], inv_p[p[y]]);
      swap(p[x], p[y]);
      long long sx = compute(x), sy = compute(y);
      for (long long r = 0; r < S; r++) {
        reach[x] = sx, reach[y] = sy;
        x = inv_p[x], y = inv_p[y];
        sx = inv_p[sx], sy = inv_p[sy];
      }
    } else {
      long long x, k;
      cin >> x >> k;
      x--;
      while (k >= S) {
        x = reach[x];
        k -= S;
      }
      while (k--) {
        x = p[x];
      }
      cout << x + 1 << '\n';
    }
  }
}
int32_t main() {
  cin.tie(0)->sync_with_stdio(0);
  long long t_c = 1;
  while (t_c--) testCase();
}

```