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 lvls = 62;
long long shift[lvls];
int get_lvl(long long x) {
  int ans = 0;
  while (x > 0) {
    x /= 2;
    ans++;
  }
  return ans - 1;
}
void add_shift(int lvl, long long k, bool rec = false) {
  if (lvl == lvls) return;
  long long lvl_max = (1LL << lvl);
  if (k >= 0)
    k %= lvl_max;
  else {
    long long ak = abs(k) / lvl_max * lvl_max;
    k = (k + ak + lvl_max) % lvl_max;
  }
  shift[lvl] += k;
  shift[lvl] %= lvl_max;
  if (rec) add_shift(lvl + 1, k * 2, rec);
}
long long init_pos(int lvl, long long x) {
  long long lvl_min_el = (1LL << lvl);
  return x - lvl_min_el;
}
long long init_el(int lvl, long long pos) {
  long long lvl_min_el = (1LL << lvl);
  return lvl_min_el + pos;
}
void print_up(int lvl, long long pos) {
  long long ipos = (pos + (1LL << lvl) - shift[lvl]) % (1LL << lvl);
  cout << init_el(lvl, ipos) << ' ';
  if (lvl == 0) {
    cout << endl;
    return;
  }
  print_up(lvl - 1, pos / 2);
}
int main() {
  ios_base::sync_with_stdio(false);
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 3) {
      long long x;
      cin >> x;
      int lvl = get_lvl(x);
      long long pos = init_pos(lvl, x);
      pos = (pos + shift[lvl]) % (1LL << lvl);
      print_up(lvl, pos);
    } else {
      long long x, k;
      cin >> x >> k;
      int lvl = get_lvl(x);
      add_shift(lvl, k, t == 2);
    }
  }
}