#include <bits/stdc++.h>
using namespace std;
long long shiftLevel[62 + 1];
long long Q, T;
long long first, K;
long long tp[62];
long long getLevel(long long first) {
  return upper_bound(tp, tp + 62 + 1, first) - tp;
}
void shiftRight(long long l, long long v) {
  shiftLevel[l] += v;
  shiftLevel[l] %= tp[l - 1];
  shiftLevel[l] += tp[l - 1];
  shiftLevel[l] %= tp[l - 1];
}
long long getValue(long long node, long long l) {
  return (node - shiftLevel[l] + tp[l - 1]) % tp[l - 1];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  tp[0] = 1;
  for (long long i = 1; i <= 62; i++) tp[i] = tp[i - 1] << 1;
  cin >> Q;
  while (Q--) {
    cin >> T;
    if (T == 3) {
      cin >> first;
      long long l = getLevel(first);
      long long node = (first + shiftLevel[l] + tp[l - 1]) % tp[l - 1];
      for (long long i = l; i >= 1; i--) {
        cout << tp[i - 1] + getValue(node, i) << " ";
        node >>= 1;
      }
      cout << "\n";
    } else {
      cin >> first >> K;
      long long l = getLevel(first);
      K = K % tp[l - 1];
      if (T == 1) {
        shiftRight(l, K);
      } else {
        for (long long i = l; i <= 62; i++) {
          shiftRight(i, K);
          K <<= 1;
        }
      }
    }
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 shiftLevel[62 + 1];
long long Q, T;
long long first, K;
long long tp[62];
long long getLevel(long long first) {
  return upper_bound(tp, tp + 62 + 1, first) - tp;
}
void shiftRight(long long l, long long v) {
  shiftLevel[l] += v;
  shiftLevel[l] %= tp[l - 1];
  shiftLevel[l] += tp[l - 1];
  shiftLevel[l] %= tp[l - 1];
}
long long getValue(long long node, long long l) {
  return (node - shiftLevel[l] + tp[l - 1]) % tp[l - 1];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  tp[0] = 1;
  for (long long i = 1; i <= 62; i++) tp[i] = tp[i - 1] << 1;
  cin >> Q;
  while (Q--) {
    cin >> T;
    if (T == 3) {
      cin >> first;
      long long l = getLevel(first);
      long long node = (first + shiftLevel[l] + tp[l - 1]) % tp[l - 1];
      for (long long i = l; i >= 1; i--) {
        cout << tp[i - 1] + getValue(node, i) << " ";
        node >>= 1;
      }
      cout << "\n";
    } else {
      cin >> first >> K;
      long long l = getLevel(first);
      K = K % tp[l - 1];
      if (T == 1) {
        shiftRight(l, K);
      } else {
        for (long long i = l; i <= 62; i++) {
          shiftRight(i, K);
          K <<= 1;
        }
      }
    }
  }
}

```