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 int st[70];
long long int no[70];
vector<long long int> vtr;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1, y;
  long long int pos, lvl, x, k;
  long long int i;
  const long long int M = 1e18;
  while (1) {
    vtr.push_back(t);
    t *= 2;
    if (t > M || t < 0) break;
  }
  t = 1;
  for (i = 0; i < vtr.size(); i++) {
    st[i] = 0;
    no[i] = vtr[i];
  }
  int q;
  cin >> q;
  int type;
  while (q--) {
    cin >> type;
    if (type == 3) {
      cin >> x;
      lvl = upper_bound(vtr.begin(), vtr.end(), x) - vtr.begin();
      lvl--;
      y = vtr[lvl];
      pos = (st[lvl] + x - y) % (no[lvl]);
      pos = (pos + no[lvl]) % (no[lvl]);
      while (lvl >= 0) {
        if (pos >= st[lvl])
          cout << vtr[lvl] + pos - st[lvl] << " ";
        else
          cout << 2 * vtr[lvl] - st[lvl] + pos << " ";
        lvl--;
        pos /= 2;
      }
      cout << "\n";
    }
    if (type == 1) {
      cin >> x >> k;
      lvl = upper_bound(vtr.begin(), vtr.end(), x) - vtr.begin();
      lvl--;
      st[lvl] = (st[lvl] + k) % no[lvl];
      st[lvl] = (st[lvl] + no[lvl]) % (no[lvl]);
    }
    if (type == 2) {
      cin >> x >> k;
      t = 1;
      lvl = upper_bound(vtr.begin(), vtr.end(), x) - vtr.begin();
      lvl--;
      for (i = lvl; i < vtr.size(); i++) {
        st[i] = (st[i] + k) % (no[i]);
        st[i] = (st[i] + no[i]) % (no[i]);
        k *= 2;
      }
    }
  }
  return 0;
}