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 main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int Q;
  cin >> Q;
  vector<long long> shift(64, 0);
  for (int qq = 0; qq < Q; ++qq) {
    int type;
    long long x;
    cin >> type;
    cin >> x;
    int level = 64 - __builtin_clzll(x);
    if (type < 3) {
      long long k;
      cin >> k;
      if (type == 1) {
        shift[level] =
            (shift[level] + k + (1LL << (level - 1))) % (1LL << (level - 1));
      } else {
        int lvl = level;
        long long amt = k;
        for (; lvl < 61; amt <<= 1, lvl++) {
          shift[lvl] =
              (shift[lvl] + amt + (1LL << (lvl - 1))) % (1LL << (lvl - 1));
        }
      }
    } else {
      int lev = level;
      long long pos =
          (x + shift[lev]) % (1LL << (lev - 1)) + (1LL << (lev - 1));
      while (pos > 0) {
        printf("%I64d ",
               (pos - shift[lev]) % (1LL << (lev - 1)) + (1LL << (lev - 1)));
        pos >>= 1;
        --lev;
      }
      printf("\n");
    }
  }
  return 0;
}