#include <bits/stdc++.h>
using namespace std;
mt19937_64 mrand(chrono::steady_clock::now().time_since_epoch().count());
const int N = 1e5 + 5;
int ch[N][2], rd[N], sz[N], val[N], a[N], fa[N], vis[N], tot = 0;
void pushup(int rt) { sz[rt] = 1 + sz[ch[rt][0]] + sz[ch[rt][1]]; }
void split(int rt, int k, int &x, int &y) {
  if (!rt) {
    x = y = 0;
    return;
  }
  if (sz[ch[rt][0]] + 1 <= k) {
    if (ch[rt][1]) fa[ch[rt][1]] = 0;
    x = rt, split(ch[rt][1], k - sz[ch[rt][0]] - 1, ch[rt][1], y);
    if (ch[rt][1]) fa[ch[rt][1]] = rt;
  } else {
    if (ch[rt][0]) fa[ch[rt][0]] = 0;
    y = rt, split(ch[rt][0], k, x, ch[rt][0]);
    if (ch[rt][0]) fa[ch[rt][0]] = rt;
  }
  pushup(rt);
}
int merge(int x, int y) {
  if (!x || !y) return x + y;
  if (rd[x] < rd[y]) {
    ch[x][1] = merge(ch[x][1], y);
    if (ch[x][1]) fa[ch[x][1]] = x;
    pushup(x);
    return x;
  } else {
    ch[y][0] = merge(x, ch[y][0]);
    if (ch[y][0]) fa[ch[y][0]] = y;
    pushup(y);
    return y;
  }
}
pair<int, int> get(int x) {
  int p = x, pos = sz[ch[x][0]];
  while (fa[p]) {
    if (p == ch[fa[p]][1]) pos += sz[ch[fa[p]][0]] + 1;
    p = fa[p];
  }
  return {p, pos};
}
int kth(int p, int k) {
  while (k) {
    if (k <= sz[ch[p][0]]) {
      p = ch[p][0];
    } else if (k == sz[ch[p][0]] + 1) {
      return p;
    } else {
      k -= sz[ch[p][0]] + 1;
      p = ch[p][1];
    }
  }
  return -1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; ++i)
    cin >> a[i], rd[i] = mrand() % INT_MAX, sz[i] = 1;
  for (int i = 1; i <= n; ++i) {
    if (vis[i]) continue;
    int p = i, rt = 0;
    while (!vis[p]) {
      vis[p] = 1;
      rt = merge(rt, p);
      p = a[p];
    }
  }
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int x, y;
      cin >> x >> y;
      auto [rtx, px] = get(x);
      auto [rty, py] = get(y);
      if (rtx == rty) {
        if (px > py) {
          swap(px, py);
          swap(x, y);
        }
        int l, m, r;
        split(rtx, px + 1, l, r);
        split(r, py - px, m, r);
        merge(l, r);
      } else {
        int lx, rx, ly, ry;
        split(rtx, px + 1, lx, rx);
        split(rty, py + 1, ly, ry);
        merge(merge(lx, ry), merge(ly, rx));
      }
    } else {
      int x, k;
      cin >> x >> k;
      auto [rt, pos] = get(x);
      pos = (pos + k) % sz[rt];
      cout << kth(rt, pos + 1) << '\n';
    }
  }
}

### Prompt

Please formulate 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;
mt19937_64 mrand(chrono::steady_clock::now().time_since_epoch().count());
const int N = 1e5 + 5;
int ch[N][2], rd[N], sz[N], val[N], a[N], fa[N], vis[N], tot = 0;
void pushup(int rt) { sz[rt] = 1 + sz[ch[rt][0]] + sz[ch[rt][1]]; }
void split(int rt, int k, int &x, int &y) {
  if (!rt) {
    x = y = 0;
    return;
  }
  if (sz[ch[rt][0]] + 1 <= k) {
    if (ch[rt][1]) fa[ch[rt][1]] = 0;
    x = rt, split(ch[rt][1], k - sz[ch[rt][0]] - 1, ch[rt][1], y);
    if (ch[rt][1]) fa[ch[rt][1]] = rt;
  } else {
    if (ch[rt][0]) fa[ch[rt][0]] = 0;
    y = rt, split(ch[rt][0], k, x, ch[rt][0]);
    if (ch[rt][0]) fa[ch[rt][0]] = rt;
  }
  pushup(rt);
}
int merge(int x, int y) {
  if (!x || !y) return x + y;
  if (rd[x] < rd[y]) {
    ch[x][1] = merge(ch[x][1], y);
    if (ch[x][1]) fa[ch[x][1]] = x;
    pushup(x);
    return x;
  } else {
    ch[y][0] = merge(x, ch[y][0]);
    if (ch[y][0]) fa[ch[y][0]] = y;
    pushup(y);
    return y;
  }
}
pair<int, int> get(int x) {
  int p = x, pos = sz[ch[x][0]];
  while (fa[p]) {
    if (p == ch[fa[p]][1]) pos += sz[ch[fa[p]][0]] + 1;
    p = fa[p];
  }
  return {p, pos};
}
int kth(int p, int k) {
  while (k) {
    if (k <= sz[ch[p][0]]) {
      p = ch[p][0];
    } else if (k == sz[ch[p][0]] + 1) {
      return p;
    } else {
      k -= sz[ch[p][0]] + 1;
      p = ch[p][1];
    }
  }
  return -1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; ++i)
    cin >> a[i], rd[i] = mrand() % INT_MAX, sz[i] = 1;
  for (int i = 1; i <= n; ++i) {
    if (vis[i]) continue;
    int p = i, rt = 0;
    while (!vis[p]) {
      vis[p] = 1;
      rt = merge(rt, p);
      p = a[p];
    }
  }
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int x, y;
      cin >> x >> y;
      auto [rtx, px] = get(x);
      auto [rty, py] = get(y);
      if (rtx == rty) {
        if (px > py) {
          swap(px, py);
          swap(x, y);
        }
        int l, m, r;
        split(rtx, px + 1, l, r);
        split(r, py - px, m, r);
        merge(l, r);
      } else {
        int lx, rx, ly, ry;
        split(rtx, px + 1, lx, rx);
        split(rty, py + 1, ly, ry);
        merge(merge(lx, ry), merge(ly, rx));
      }
    } else {
      int x, k;
      cin >> x >> k;
      auto [rt, pos] = get(x);
      pos = (pos + k) % sz[rt];
      cout << kth(rt, pos + 1) << '\n';
    }
  }
}

```