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 q, t;
long long x, k, add[64];
long long transform(long long u) {
  int bs = 63 - __builtin_clzll(u);
  long long md = 1LL << bs;
  u -= md;
  return (u + add[bs]) % md + md;
}
long long value(long long u) {
  int bs = 63 - __builtin_clzll(u);
  long long md = 1LL << bs;
  u -= md;
  return (u - add[bs] + md) % md + md;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> q;
  while (q--) {
    cin >> t >> x;
    if (t <= 2) {
      int bs = 63 - __builtin_clzll(x);
      long long md = 1LL << bs;
      cin >> k;
      ((k %= md) += md) %= md;
      for (int i = bs; i <= (t == 1 ? bs : 60); i++) {
        (add[i] += k) %= md;
        k *= 2;
        md *= 2;
      }
    } else {
      long long ori = transform(x);
      while (ori != 0) {
        cout << value(ori) << " ";
        ori /= 2;
      }
      cout << '\n';
    }
  }
}