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;
int q, op;
long long x, y, sh[64];
inline void shift(int fl, long long val) {
  register long long mod = 1LL << fl;
  val %= mod;
  sh[fl] = (sh[fl] + val + mod) % mod;
}
int main() {
  scanf("%d", &q);
  while (q--) {
    int be = 0;
    scanf("%d%lld", &op, &x);
    for (be = 0; x >> be; be++)
      ;
    be--;
    if (op < 3) {
      scanf("%lld", &y);
      if (op > 1)
        shift(be, y);
      else
        shift(be, y), shift(be + 1, -y * 2);
    } else {
      for (; be >= 0; be--) {
        printf("%lld ", x);
        x += sh[be];
        if (x >= (1LL << be + 1)) x -= (1LL << be);
        x >>= 1;
      }
      puts("");
    }
  }
  return 0;
}