#include <bits/stdc++.h>
using namespace std;
long long arr[100], x, k;
long long solve(int level, long long pos) {
  long long v = (1LL << level);
  return (pos - arr[level] + v) % v + v;
}
int getlevel(long long v) {
  for (int i = 0; i < 62; i++) {
    if ((1LL << i) <= v && v < (1LL << (i + 1))) {
      return i;
    }
  }
}
long long getpos(long long x, int level) {
  long long v = (1LL << level);
  return (x + v) % v;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  ;
  int Q;
  cin >> Q;
  while (Q--) {
    int type;
    cin >> type;
    if (type < 3) {
      cin >> x >> k;
      int level = getlevel(x);
      long long v = (1LL << level);
      k = (k + v) % v;
      arr[level] = (arr[level] + k) % v;
      if (type > 1) {
        while (level++ < 62) {
          k = 2LL * k, v = 2LL * v;
          arr[level] = (arr[level] + k) % v;
        }
      }
    } else {
      cin >> x;
      int level = getlevel(x);
      long long v = (1LL << level);
      long long now = v + (arr[level] + getpos(x, level)) % v;
      while (level >= 0) {
        cout << solve(level, getpos(now, level)) << " ";
        now = now / 2LL, level--;
      }
      cout << endl;
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution 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>
using namespace std;
long long arr[100], x, k;
long long solve(int level, long long pos) {
  long long v = (1LL << level);
  return (pos - arr[level] + v) % v + v;
}
int getlevel(long long v) {
  for (int i = 0; i < 62; i++) {
    if ((1LL << i) <= v && v < (1LL << (i + 1))) {
      return i;
    }
  }
}
long long getpos(long long x, int level) {
  long long v = (1LL << level);
  return (x + v) % v;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  ;
  int Q;
  cin >> Q;
  while (Q--) {
    int type;
    cin >> type;
    if (type < 3) {
      cin >> x >> k;
      int level = getlevel(x);
      long long v = (1LL << level);
      k = (k + v) % v;
      arr[level] = (arr[level] + k) % v;
      if (type > 1) {
        while (level++ < 62) {
          k = 2LL * k, v = 2LL * v;
          arr[level] = (arr[level] + k) % v;
        }
      }
    } else {
      cin >> x;
      int level = getlevel(x);
      long long v = (1LL << level);
      long long now = v + (arr[level] + getpos(x, level)) % v;
      while (level >= 0) {
        cout << solve(level, getpos(now, level)) << " ";
        now = now / 2LL, level--;
      }
      cout << endl;
    }
  }
  return 0;
}

```