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 Q, op;
long long tam[60 + 5];
long long shift[60 + 5];
long long nivel(long long x) {
  long long tmp = 0;
  while (x) {
    tmp++;
    x /= 2;
  }
  return tmp;
}
long long val(long long x, long long n) {
  long long tmp = x - shift[n];
  if (tmp >= tam[n]) return tmp;
  tmp += tam[n];
  return tmp;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  tam[1] = 1;
  for (long long i = 2; i <= 60 + 2; i++) tam[i] = tam[i - 1] * 2;
  cin >> Q;
  while (Q--) {
    cin >> op;
    if (op == 1) {
      long long x, k;
      cin >> x >> k;
      long long n = nivel(x);
      shift[n] += (k % tam[n]);
      while (shift[n] < 0) shift[n] += tam[n];
      shift[n] %= tam[n];
    } else if (op == 2) {
      long long x, k;
      cin >> x >> k;
      long long n = nivel(x);
      long long pot = 1;
      while (n <= 60) {
        shift[n] += (((k % tam[n]) * pot) % tam[n]);
        while (shift[n] < 0) shift[n] += tam[n];
        shift[n] %= tam[n];
        pot *= 2;
        n++;
      }
    } else {
      long long x;
      cin >> x;
      long long n = nivel(x);
      long long pos = (x + shift[n]);
      if (pos >= tam[n + 1]) pos -= tam[n];
      while (n >= 1) {
        cout << val(pos, n) << " ";
        pos /= 2;
        n--;
      }
      cout << "\n";
    }
  }
  return 0;
}