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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long a[65];
int get_msb_index(long long X) {
  for (int i = 62; i >= 0; i--)
    if (X & (1LL << i)) return i;
  return -1;
}
int main() {
  int Q;
  cin >> Q;
  while (Q--) {
    int t;
    long long x, k;
    cin >> t >> x;
    int ix = get_msb_index(x);
    long long L = 1LL << ix, R = (L << 1LL) - 1, range = R - L + 1;
    if (t == 1) {
      cin >> k;
      k %= range;
      if (k < 0) k += range;
      a[ix] = (a[ix] + k) % range;
    } else if (t == 2) {
      cin >> k;
      long long _range = range;
      for (int i = ix; i < 63; i++, _range <<= 1) {
        long long _k = k << (i - ix);
        _k %= _range;
        if (_k < 0) _k += _range;
        a[i] = (a[i] + _k) % _range;
      }
    } else {
      cout << x << " ";
      x = (x + a[ix] > R ? L + (x + a[ix]) % R - 1 : x + a[ix]) / 2;
      ix--;
      L /= 2;
      R /= 2;
      range /= 2;
      while (x) {
        long long cmp = (range - a[ix]) % range;
        long long _x = (x + cmp > R ? L + (x + cmp) % R - 1 : x + cmp);
        cout << _x << " ";
        x /= 2;
        ix--;
        L /= 2;
        R /= 2;
        range /= 2;
      }
      cout << endl;
    }
  }
  return 0;
}