#include <bits/stdc++.h>
using namespace std;
int calc(long long x) {
  int res = 0;
  while (x) {
    res++;
    x >>= 1;
  }
  return res;
}
long long tag[65], pre[65];
int main() {
  ios::sync_with_stdio(false);
  pre[0] = 1;
  for (int i = 1; i <= 60; i++) pre[i] = pre[i - 1] * 2;
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      int le = calc(x);
      tag[le] = (tag[le] + k % pre[le - 1] + pre[le - 1]) % pre[le - 1];
    } else if (t == 2) {
      long long x, k;
      cin >> x >> k;
      int le = calc(x);
      k = (k % pre[le - 1] + pre[le - 1]) % pre[le - 1];
      for (int i = le; i <= 60; i++) {
        tag[i] = (tag[i] + k % pre[i - 1] + pre[i - 1]) % pre[i - 1];
        k <<= 1;
      }
    } else {
      long long x;
      cin >> x;
      int le = calc(x);
      while (x) {
        cout << x << " ";
        x = pre[le - 1] + (x - pre[le - 1] + tag[le]) % pre[le - 1];
        x >>= 1;
        if (!x) break;
        le--;
        x = pre[le - 1] + (x - tag[le]) % pre[le - 1];
      }
      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;
int calc(long long x) {
  int res = 0;
  while (x) {
    res++;
    x >>= 1;
  }
  return res;
}
long long tag[65], pre[65];
int main() {
  ios::sync_with_stdio(false);
  pre[0] = 1;
  for (int i = 1; i <= 60; i++) pre[i] = pre[i - 1] * 2;
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      int le = calc(x);
      tag[le] = (tag[le] + k % pre[le - 1] + pre[le - 1]) % pre[le - 1];
    } else if (t == 2) {
      long long x, k;
      cin >> x >> k;
      int le = calc(x);
      k = (k % pre[le - 1] + pre[le - 1]) % pre[le - 1];
      for (int i = le; i <= 60; i++) {
        tag[i] = (tag[i] + k % pre[i - 1] + pre[i - 1]) % pre[i - 1];
        k <<= 1;
      }
    } else {
      long long x;
      cin >> x;
      int le = calc(x);
      while (x) {
        cout << x << " ";
        x = pre[le - 1] + (x - pre[le - 1] + tag[le]) % pre[le - 1];
        x >>= 1;
        if (!x) break;
        le--;
        x = pre[le - 1] + (x - tag[le]) % pre[le - 1];
      }
      cout << endl;
    }
  }
  return 0;
}

```