#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e6 + 50;
template <typename A, typename B>
istream& operator>>(istream& o, pair<A, B>& a) {
  return o >> a.first >> a.second;
}
template <typename A, typename B>
ostream& operator<<(ostream& o, pair<A, B> a) {
  return o << '(' << a.first << ", " << a.second << ')';
}
namespace io {
const long long SIZE = 1e7 + 10;
char inbuff[SIZE];
char *l, *r;
inline void init() {
  l = inbuff;
  r = inbuff + fread(inbuff, 1, SIZE, stdin);
}
inline char gc() {
  if (l == r) init();
  return (l != r) ? *(l++) : EOF;
}
void R(long long& x) {
  bool neg = 0;
  x = 0;
  char ch = gc();
  while (!isdigit(ch)) {
    if (ch == '-') neg = 1;
    ch = gc();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = gc();
  if (neg) x = -x;
}
}  // namespace io
using io::R;
void solve() {
  long long n, q, sz = 300, tot = 0;
  R(n), R(q);
  vector<long long> a(n), id(n), jp(n), r(n);
  for (long long& o : a) R(o), o--;
  for (long long i = 0; i <= n - 1; i++) {
    if (!id[a[i]]) {
      tot++;
      long long t = a[i];
      while (!id[t]) {
        id[t] = tot;
        r[a[t]] = t;
        t = a[t];
      }
    }
  }
  for (long long i = 0; i <= n - 1; i++) {
    long long t = i;
    for (long long j = 1; j <= sz; j++) t = a[t];
    jp[i] = t;
  }
  while (q--) {
    long long o, x, y;
    R(o), R(x), R(y);
    x--;
    if (o == 1) {
      y--;
      long long tx = x, gpx = x, ty = y, gpy = y;
      r[a[x]] = y;
      r[a[y]] = x;
      swap(a[x], a[y]);
      for (long long i = 1; i <= sz; i++) gpx = a[gpx], gpy = a[gpy];
      jp[x] = gpx, jp[y] = gpy;
      for (long long i = 1; i <= sz; i++) {
        tx = r[tx];
        ty = r[ty];
        gpx = r[gpx];
        gpy = r[gpy];
        jp[tx] = gpx;
        jp[ty] = gpy;
      }
    } else {
      long long t = x;
      for (long long i = 1; i <= y / sz; i++) t = jp[t];
      for (long long i = 1; i <= y - y / sz * sz; i++) t = a[t];
      printf("%lld\n", t + 1);
    }
  }
}
void input() {}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  ;
  long long tt = 1;
  while (tt--) {
    input();
    solve();
  }
}

### 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 long long maxn = 1e6 + 50;
template <typename A, typename B>
istream& operator>>(istream& o, pair<A, B>& a) {
  return o >> a.first >> a.second;
}
template <typename A, typename B>
ostream& operator<<(ostream& o, pair<A, B> a) {
  return o << '(' << a.first << ", " << a.second << ')';
}
namespace io {
const long long SIZE = 1e7 + 10;
char inbuff[SIZE];
char *l, *r;
inline void init() {
  l = inbuff;
  r = inbuff + fread(inbuff, 1, SIZE, stdin);
}
inline char gc() {
  if (l == r) init();
  return (l != r) ? *(l++) : EOF;
}
void R(long long& x) {
  bool neg = 0;
  x = 0;
  char ch = gc();
  while (!isdigit(ch)) {
    if (ch == '-') neg = 1;
    ch = gc();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = gc();
  if (neg) x = -x;
}
}  // namespace io
using io::R;
void solve() {
  long long n, q, sz = 300, tot = 0;
  R(n), R(q);
  vector<long long> a(n), id(n), jp(n), r(n);
  for (long long& o : a) R(o), o--;
  for (long long i = 0; i <= n - 1; i++) {
    if (!id[a[i]]) {
      tot++;
      long long t = a[i];
      while (!id[t]) {
        id[t] = tot;
        r[a[t]] = t;
        t = a[t];
      }
    }
  }
  for (long long i = 0; i <= n - 1; i++) {
    long long t = i;
    for (long long j = 1; j <= sz; j++) t = a[t];
    jp[i] = t;
  }
  while (q--) {
    long long o, x, y;
    R(o), R(x), R(y);
    x--;
    if (o == 1) {
      y--;
      long long tx = x, gpx = x, ty = y, gpy = y;
      r[a[x]] = y;
      r[a[y]] = x;
      swap(a[x], a[y]);
      for (long long i = 1; i <= sz; i++) gpx = a[gpx], gpy = a[gpy];
      jp[x] = gpx, jp[y] = gpy;
      for (long long i = 1; i <= sz; i++) {
        tx = r[tx];
        ty = r[ty];
        gpx = r[gpx];
        gpy = r[gpy];
        jp[tx] = gpx;
        jp[ty] = gpy;
      }
    } else {
      long long t = x;
      for (long long i = 1; i <= y / sz; i++) t = jp[t];
      for (long long i = 1; i <= y - y / sz * sz; i++) t = a[t];
      printf("%lld\n", t + 1);
    }
  }
}
void input() {}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  ;
  long long tt = 1;
  while (tt--) {
    input();
    solve();
  }
}

```