#include <bits/stdc++.h>
using namespace std;
long long v[100], n[100];
void moduluj(long long& co, long long m) {
  if (co >= 0)
    co = co % m;
  else
    co = m - abs(co) % m;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long q;
  long long a, b, c;
  long long gdzie;
  cin >> q;
  while (q--) {
    cin >> a >> b;
    if (a == 3) {
      c = b;
      gdzie = 0;
      long long x = 1;
      while (c /= 2) {
        gdzie++;
        x *= 2;
      }
      cout << b << " ";
      b = (b - v[gdzie] + x + n[gdzie]) % x + x;
      b /= 2;
      x /= 2;
      gdzie--;
      while (b) {
        cout << (b + v[gdzie]) % x + x << " ";
        b = (b + n[gdzie]) % x + x;
        b /= 2;
        x /= 2;
        gdzie--;
      }
      cout << "\n";
    } else {
      cin >> c;
      gdzie = 0;
      long long x = 1;
      while (b /= 2) {
        gdzie++;
        x *= 2;
      }
      if (a == 1) {
        v[gdzie] -= c;
        moduluj(v[gdzie], x);
      } else {
        n[gdzie] += c;
        moduluj(n[gdzie], x);
      }
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
using namespace std;
long long v[100], n[100];
void moduluj(long long& co, long long m) {
  if (co >= 0)
    co = co % m;
  else
    co = m - abs(co) % m;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long q;
  long long a, b, c;
  long long gdzie;
  cin >> q;
  while (q--) {
    cin >> a >> b;
    if (a == 3) {
      c = b;
      gdzie = 0;
      long long x = 1;
      while (c /= 2) {
        gdzie++;
        x *= 2;
      }
      cout << b << " ";
      b = (b - v[gdzie] + x + n[gdzie]) % x + x;
      b /= 2;
      x /= 2;
      gdzie--;
      while (b) {
        cout << (b + v[gdzie]) % x + x << " ";
        b = (b + n[gdzie]) % x + x;
        b /= 2;
        x /= 2;
        gdzie--;
      }
      cout << "\n";
    } else {
      cin >> c;
      gdzie = 0;
      long long x = 1;
      while (b /= 2) {
        gdzie++;
        x *= 2;
      }
      if (a == 1) {
        v[gdzie] -= c;
        moduluj(v[gdzie], x);
      } else {
        n[gdzie] += c;
        moduluj(n[gdzie], x);
      }
    }
  }
  return 0;
}

```