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 int modu(long long int a, long long int x) {
  if (a % x >= 0)
    return a % x;
  else
    return a % x + x;
}
int main(int argc, char const *argv[]) {
  long long int p = 1;
  long long int pow2[61];
  for (int i = 0; i < 61; ++i) {
    pow2[i] = p;
    p <<= 1;
  }
  long long int q;
  cin >> q;
  long long int shift[61] = {0};
  long long int t;
  long long int x, k;
  long long int lev;
  for (long long int i = 0; i < q; ++i) {
    cin >> t;
    switch (t) {
      case 1:
        cin >> x >> k;
        lev = (long long int)log2(x);
        shift[lev] = modu((shift[lev] + k % pow2[lev]), (pow2[lev]));
        break;
      case 2:
        cin >> x >> k;
        lev = (long long int)log2(x);
        for (int i = lev; i < 61; ++i) {
          shift[i] = modu((shift[i] + k % pow2[i]), (pow2[i]));
          k <<= 1;
        }
        break;
      case 3:
        cin >> x;
        lev = (long long int)log2(x);
        x = modu((x + shift[lev]), (pow2[lev])) + pow2[lev];
        while (x) {
          cout << modu((x - shift[lev]), (pow2[lev])) + pow2[lev] << " ";
          x >>= 1;
          lev--;
        }
        cout << endl;
        break;
      default:
        break;
    }
  }
}