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>
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> q;
  long long moved[60] = {0};
  for (int t = 0; t < q; t++) {
    int type;
    cin >> type;
    switch (type) {
      case 1: {
        long long x, d;
        cin >> x >> d;
        int p2 = 63 - __builtin_clzll(x);
        d %= (long long)1 << ((int)log2(x));
        moved[p2] = (moved[p2] + d) % ((long long)1 << p2);
        break;
      }
      case 2: {
        long long x, d;
        cin >> x >> d;
        d %= (long long)1 << ((int)log2(x));
        for (int p2 = 63 - __builtin_clzll(x); p2 < 60; p2++, d *= 2)
          moved[p2] = (moved[p2] + d) % ((long long)1 << p2);
        break;
      }
      case 3: {
        long long x;
        cin >> x;
        int p2 = 63 - __builtin_clzll(x);
        cout << x << ' ';
        x = ((long long)1 << p2) + ((x + moved[p2]) % ((long long)1 << p2));
        x /= 2;
        p2--;
        while (x) {
          cout << ((long long)1 << p2) +
                      ((x - moved[p2]) % ((long long)1 << p2))
               << ' ';
          ;
          p2--;
          x /= 2;
        }
        cout << '\n';
        break;
      }
    }
  }
  return 0;
}
