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>
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class FullBinaryTreeQueries {
	static long maxNode = (long) Math.pow(2, 60) - 1;

	public static void main(String[] args) throws NumberFormatException, IOException {
		int depth = log2(maxNode);
		long[] shift = new long[depth + 2];
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int Q = Integer.parseInt(br.readLine());
		for (int i = 0; i < Q; i++) {
			long[] in = Arrays.stream(br.readLine().split(" ")).mapToLong(x -> Long.parseLong(x)).toArray();
			int nodeDepth = log2(in[1]);
			if (in[0] == 1 || in[0] == 2) {
				long rot = ((in[2] % (1l << nodeDepth)) + (1l << nodeDepth)) % (1l << nodeDepth);
				shift[nodeDepth] = (shift[nodeDepth] + rot) % (1l << nodeDepth);
				if (in[0] == 2) {
					while (nodeDepth < depth) {
						nodeDepth++;
						rot *= 2;
						shift[nodeDepth] = (shift[nodeDepth] + rot) % (1l << nodeDepth);
					}
				}
			} else {
				long loc = in[1];
				long curLoc = (loc + shift[nodeDepth]) % (1l << nodeDepth) + (1l << nodeDepth);
				StringBuilder ans = new StringBuilder("" + loc);
				curLoc /= 2;
				while (curLoc != 0) {
					nodeDepth--;
					loc = ((1l << nodeDepth) + curLoc - shift[nodeDepth]) % (1l << nodeDepth) + (1l << nodeDepth);
					ans.append(" " + loc);
					curLoc /= 2;
				}
				System.out.println(ans);
			}
		}
	}

	public static int log2(long n) {
		return 63 - Long.numberOfLeadingZeros(n);
	}
}
