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 main() {
  long long shifts[61] = {0};
  int q;
  cin >> q;
  while (q--) {
    int t;
    long long x, k;
    cin >> t >> x;
    long long l = log2(x);
    if (t == 3) {
      long long v =
          (1LL << l) +
          ((x % (1LL << l) + (1LL << l)) % (1LL << l) + shifts[l]) % (1LL << l);
      while (v) {
        l = log2(v);
        long long z = (1LL << l) +
                      ((v % (1LL << l) - shifts[l]) % (1LL << l) + (1LL << l)) %
                          (1LL << l);
        v /= 2;
        cout << z << " ";
      }
      cout << endl;
    } else {
      cin >> k;
      if (t == 1)
        shifts[l] = ((shifts[l] + k) % (1LL << l) + (1LL << l)) % (1LL << l);
      else {
        for (long long i = l; i <= 60; i++)
          shifts[i] =
              (shifts[i] + ((1LL << (i - l)) * k) % (1LL << i) + (1LL << i)) %
              (1LL << i);
      }
    }
  }
  return 0;
}