#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
template <class T>
using V = vector<T>;
const ll inf = (1e18);
const ll mod = 1000000007;
const vector<int> dy = {-1, 0, 1, 0}, dx = {0, -1, 0, 1};
ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
ll LCM(ll c, ll d) { return c / GCD(c, d) * d; }
struct __INIT {
  __INIT() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
  }
} __init;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
void debag(const vector<T> &a) {
  cerr << "debag :";
  for (auto v : a) cerr << v << " ";
  cerr << "\n";
}
template <class T>
void print(const vector<T> &a) {
  for (auto v : a) cout << v << " ";
  cout << "\n";
}
void solve() {
  int n, q;
  cin >> n >> q;
  int B = ll(sqrt(double(n))) + 10;
  V<int> f(n), fb(n), inv(n);
  iota((fb).begin(), (fb).end(), 0);
  for (int i = 0; i < n; i++) {
    cin >> f[i];
    f[i]--;
  }
  for (int i = 0; i < n; i++) {
    inv[f[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < B; j++) {
      fb[i] = f[fb[i]];
    }
  }
  V<bool> used(n, true);
  for (int i = 0; i < q; i++) {
    int t, x, y;
    cin >> t >> x >> y;
    --x;
    --y;
    if (t == 1) {
      V<int> tmp;
      int cp = x;
      for (int j = 0; j <= B; j++) {
        tmp.emplace_back(cp);
        used[cp] = false;
        cp = inv[cp];
        if (cp == x) break;
      }
      cp = y;
      for (int j = 0; j <= B; j++) {
        tmp.emplace_back(cp);
        used[cp] = false;
        cp = inv[cp];
        if (cp == y) break;
      }
      swap(f[x], f[y]);
      swap(inv[f[x]], inv[f[y]]);
      for (int v : tmp) {
        if (!used[v]) {
          if (used[f[v]]) {
            fb[v] = inv[fb[f[v]]];
          } else {
            fb[v] = v;
            for (int j = 0; j < B; j++) {
              fb[v] = f[fb[v]];
            }
          }
          used[v] = true;
        }
      }
    } else {
      y++;
      int cur = x;
      while (B <= y) {
        cur = fb[cur];
        y -= B;
      }
      while (y > 0) {
        cur = f[cur];
        y--;
      }
      cout << cur + 1 << "\n";
    }
  }
}
int main() {
  int t = 1;
  while (t--) 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>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
template <class T>
using V = vector<T>;
const ll inf = (1e18);
const ll mod = 1000000007;
const vector<int> dy = {-1, 0, 1, 0}, dx = {0, -1, 0, 1};
ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
ll LCM(ll c, ll d) { return c / GCD(c, d) * d; }
struct __INIT {
  __INIT() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
  }
} __init;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
void debag(const vector<T> &a) {
  cerr << "debag :";
  for (auto v : a) cerr << v << " ";
  cerr << "\n";
}
template <class T>
void print(const vector<T> &a) {
  for (auto v : a) cout << v << " ";
  cout << "\n";
}
void solve() {
  int n, q;
  cin >> n >> q;
  int B = ll(sqrt(double(n))) + 10;
  V<int> f(n), fb(n), inv(n);
  iota((fb).begin(), (fb).end(), 0);
  for (int i = 0; i < n; i++) {
    cin >> f[i];
    f[i]--;
  }
  for (int i = 0; i < n; i++) {
    inv[f[i]] = i;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < B; j++) {
      fb[i] = f[fb[i]];
    }
  }
  V<bool> used(n, true);
  for (int i = 0; i < q; i++) {
    int t, x, y;
    cin >> t >> x >> y;
    --x;
    --y;
    if (t == 1) {
      V<int> tmp;
      int cp = x;
      for (int j = 0; j <= B; j++) {
        tmp.emplace_back(cp);
        used[cp] = false;
        cp = inv[cp];
        if (cp == x) break;
      }
      cp = y;
      for (int j = 0; j <= B; j++) {
        tmp.emplace_back(cp);
        used[cp] = false;
        cp = inv[cp];
        if (cp == y) break;
      }
      swap(f[x], f[y]);
      swap(inv[f[x]], inv[f[y]]);
      for (int v : tmp) {
        if (!used[v]) {
          if (used[f[v]]) {
            fb[v] = inv[fb[f[v]]];
          } else {
            fb[v] = v;
            for (int j = 0; j < B; j++) {
              fb[v] = f[fb[v]];
            }
          }
          used[v] = true;
        }
      }
    } else {
      y++;
      int cur = x;
      while (B <= y) {
        cur = fb[cur];
        y -= B;
      }
      while (y > 0) {
        cur = f[cur];
        y--;
      }
      cout << cur + 1 << "\n";
    }
  }
}
int main() {
  int t = 1;
  while (t--) solve();
}

```