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 JAVA):

Solution:
import java.util.*;
import java.math.*;
import java.io.*;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
public class Hello {

  
  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
 
    int q = in.nextInt();
    
    long []shift = new long [70];
 
    try {
    BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out, "UTF-8"));

    for (int i = 1;i <= q;++i) {
      int type = in.nextInt();
      long x = 0;//in.nextLong();
      long K = 0;//in.nextLong();
      
      if (type == 1 || type == 2) {
	x = in.nextLong();
	K = -in.nextLong();
	if (type == 1) {
	  int level = 63 - Long.numberOfLeadingZeros(x);//x = 1 level = 0
	  shift[level] += K;
	} else if (type == 2) {
	  int level = 63 - Long.numberOfLeadingZeros(x);//x = 1 level = 0
	  long curK = K;
	  for (int z = level;z <= 63;++z) {
	    shift[z] += curK;
	    curK = curK * 2;
	  }
	  
	}
      } else if (type == 3) {
	x = in.nextLong();
	int level = 63 - Long.numberOfLeadingZeros(x);//x = 1 level = 0
	long pos = x - shift[level];
	do {
	  out.write((1L << level) + ((pos + shift[level]) & ((1L << level) - 1)) + " ");
	  //System.out.print(' ');
	  pos >>= 1;
	} while (--level >= 0);

	out.write("\n");
      }
    }
    out.flush(); out.close();
    } catch (Exception e) {
      System.exit(0);
    }

  }
}