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;
const int NMAX = 70;
long long sval[NMAX];
int main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    long long type, x, k;
    cin >> type;
    if (type == 1) {
      cin >> x >> k;
      long long i;
      for (i = 0; (1LL << i) <= x; i++)
        ;
      i--;
      if (k < 0) k = (1LL << i) - ((-k) % (1LL << i));
      sval[i] = (sval[i] + k) % (1LL << i);
    } else if (type == 2) {
      cin >> x >> k;
      long long i;
      for (i = 0; (1LL << i) <= x; i++)
        ;
      i--;
      if (k < 0) k = (1LL << i) - ((-k) % (1LL << i));
      k %= (1LL << i);
      for (long long j = i; j <= 60; j++) {
        sval[j] = (sval[j] + k) % (1LL << j);
        k *= 2;
      }
    } else {
      cin >> x;
      long long i;
      for (i = 0; (1LL << i) <= x; i++)
        ;
      i--;
      long long node = x;
      for (long long j = i; j > 0; j--) {
        cout << node << " ";
        long long pos = node - (1LL << j);
        pos += sval[j];
        pos %= (1LL << j);
        long long nxtpos = pos / 2;
        nxtpos += (1LL << (j - 1)) - sval[j - 1];
        nxtpos %= (1LL << (j - 1));
        node = (1LL << (j - 1)) + nxtpos;
      }
      cout << node << "\n";
    }
  }
  return 0;
}