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, t[63], first, second, z;
long long lvl(long long first) {
  for (long long i = 62; i >= 0; i--) {
    if ((1ll << i) <= first) return i;
  }
}
int32_t main() {
  cin >> q;
  while (q--) {
    cin >> first;
    if (first == 1) {
      cin >> first >> second;
      long long k = lvl(first);
      t[k] += second;
      t[k] %= (1ll << k);
      t[k] += (1ll << k);
      t[k] %= (1ll << k);
    } else if (first == 2) {
      cin >> first >> second;
      long long k = lvl(first);
      second %= (1ll << k);
      for (long long i = k; i <= 62; i++) {
        t[i] += second * (1ll << (i - k));
        t[i] %= (1ll << i);
        t[i] += (1ll << i);
        t[i] %= (1ll << i);
      }
    } else {
      cin >> first;
      long long k = lvl(first);
      first += t[k];
      first %= (1ll << k);
      first += (1ll << k);
      for (long long i = k; i >= 0; i--) {
        long long p = first;
        p -= t[i];
        p %= (1ll << i);
        p += (1ll << i);
        cout << p << ' ';
        first /= 2;
      }
      cout << '\n';
    }
  }
}