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 a[100], b[100];
long long arr[100];
int main() {
  int q, t;
  long long x;
  long long k;
  long long hV = 0;
  cin >> q;
  while (q--) {
    int h;
    cin >> t >> x;
    if (t == 1) {
      h = 1;
      cin >> k;
      while (x >> h) h++;
      h--;
      a[h] += k;
      a[h] %= (1ll << h);
    }
    if (t == 2) {
      h = 1;
      cin >> k;
      while (x >> h) h++;
      h--;
      b[h] += k;
      b[h] %= (1ll << h);
    }
    if (t == 3) {
      h = 1;
      while (x >> h) h++;
      arr[0] = 0;
      for (int i = 0; i < h; ++i) {
        hV += b[i];
        hV %= (1ll << i);
        arr[i] = (hV + a[i]) % (1ll << i);
        hV <<= 1;
      }
      for (int i = h - 1; i >= 0; --i) {
        cout << x << ' ';
        hV = (1ll << i);
        x = (((x + arr[i]) % hV + hV) % hV + hV) >> 1;
        if (i) x = ((x - arr[i - 1]) % (hV / 2) + hV / 2) % (hV / 2) + hV / 2;
      }
      cout << '\n';
    }
  }
}