#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:512000000")
using namespace std;
void solve(__attribute__((unused)) bool);
void precalc();
clock_t start;
int main() {
  start = clock();
  int t = 1;
  cout.sync_with_stdio(0);
  cin.tie(0);
  precalc();
  cout.precision(10);
  cout << fixed;
  int testNum = 1;
  while (t--) {
    solve(true);
  }
  cout.flush();
  return 0;
}
void precalc() {}
template <typename T>
inline T nxt() {
  T cur;
  cin >> cur;
  return cur;
}
const long long L = 60;
long long get_level(long long x) {
  long long res = 0;
  while (x > 1) {
    x >>= 1;
    ++res;
  }
  return res;
}
void solve(__attribute__((unused)) bool read) {
  vector<long long> shift(L, 0);
  long long Q;
  cin >> Q;
  while (Q--) {
    long long type, x;
    cin >> type >> x;
    long long lev = get_level(x);
    if (type == 1) {
      long long k;
      cin >> k;
      shift[lev] = (shift[lev] + k) % (1LL << lev);
      if (shift[lev] < 0) {
        shift[lev] += (1LL << lev);
      }
    } else if (type == 2) {
      long long k;
      cin >> k;
      for (long long i = lev; i < L; ++i) {
        shift[i] = (shift[i] + k) % (1LL << i);
        if (shift[i] < 0) {
          shift[i] += (1LL << i);
        }
        k *= 2;
      }
    } else {
      long long old_pos = x - (1LL << lev);
      long long new_pos = (old_pos + shift[lev]) % (1LL << lev);
      long long cur_v = new_pos + (1LL << lev);
      while (cur_v > 1) {
        long long pos = cur_v - (1LL << lev);
        long long val = (pos - shift[lev] + (1LL << lev)) % (1LL << lev);
        cout << val + (1LL << lev) << " ";
        cur_v /= 2;
        --lev;
      }
      cout << cur_v << "\n";
    }
  }
}

### Prompt

Please create a solution in Cpp to the following problem:

You have a full binary tree having infinite levels.

Each node has an initial value. If a node has value x, then its left child has value 2·x and its right child has value 2·x + 1. 

The value of the root is 1. 

You need to answer Q queries. 

There are 3 types of queries: 

  1. Cyclically shift the values of all nodes on the same level as node with value X by K units. (The values/nodes of any other level are not affected).
  2. Cyclically shift the nodes on the same level as node with value X by K units. (The subtrees of these nodes will move along with them).
  3. Print the value of every node encountered on the simple path from the node with value X to the root.



Positive K implies right cyclic shift and negative K implies left cyclic shift. 

It is guaranteed that atleast one type 3 query is present.

Input

The first line contains a single integer Q (1 ≤ Q ≤ 105).

Then Q queries follow, one per line: 

  * Queries of type 1 and 2 have the following format: T X K (1 ≤ T ≤ 2; 1 ≤ X ≤ 1018; 0 ≤ |K| ≤ 1018), where T is type of the query.
  * Queries of type 3 have the following format: 3 X (1 ≤ X ≤ 1018).

Output

For each query of type 3, print the values of all nodes encountered in descending order.

Examples

Input

5
3 12
1 2 1
3 12
2 4 -1
3 8


Output

12 6 3 1 
12 6 2 1 
8 4 2 1 


Input

5
3 14
1 5 -3
3 14
1 3 1
3 14


Output

14 7 3 1 
14 6 3 1 
14 6 2 1 

Note

Following are the images of the first 4 levels of the tree in the first test case:

Original: 

<image>

After query 1 2 1: 

<image>

After query 2 4 -1: 

<image>

### Response

```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:512000000")
using namespace std;
void solve(__attribute__((unused)) bool);
void precalc();
clock_t start;
int main() {
  start = clock();
  int t = 1;
  cout.sync_with_stdio(0);
  cin.tie(0);
  precalc();
  cout.precision(10);
  cout << fixed;
  int testNum = 1;
  while (t--) {
    solve(true);
  }
  cout.flush();
  return 0;
}
void precalc() {}
template <typename T>
inline T nxt() {
  T cur;
  cin >> cur;
  return cur;
}
const long long L = 60;
long long get_level(long long x) {
  long long res = 0;
  while (x > 1) {
    x >>= 1;
    ++res;
  }
  return res;
}
void solve(__attribute__((unused)) bool read) {
  vector<long long> shift(L, 0);
  long long Q;
  cin >> Q;
  while (Q--) {
    long long type, x;
    cin >> type >> x;
    long long lev = get_level(x);
    if (type == 1) {
      long long k;
      cin >> k;
      shift[lev] = (shift[lev] + k) % (1LL << lev);
      if (shift[lev] < 0) {
        shift[lev] += (1LL << lev);
      }
    } else if (type == 2) {
      long long k;
      cin >> k;
      for (long long i = lev; i < L; ++i) {
        shift[i] = (shift[i] + k) % (1LL << i);
        if (shift[i] < 0) {
          shift[i] += (1LL << i);
        }
        k *= 2;
      }
    } else {
      long long old_pos = x - (1LL << lev);
      long long new_pos = (old_pos + shift[lev]) % (1LL << lev);
      long long cur_v = new_pos + (1LL << lev);
      while (cur_v > 1) {
        long long pos = cur_v - (1LL << lev);
        long long val = (pos - shift[lev] + (1LL << lev)) % (1LL << lev);
        cout << val + (1LL << lev) << " ";
        cur_v /= 2;
        --lev;
      }
      cout << cur_v << "\n";
    }
  }
}

```