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;
const long long mod = 1e9 + 7, siz = 1e6 + 5;
long long q, n, m, a[siz], rot[66];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> q;
  while (q--) {
    long long op;
    cin >> op;
    if (op <= 2) {
      long long node, k;
      cin >> node >> k;
      long long dep = log2l(node);
      long long sz = 1LL << dep;
      if (k >= 0)
        k %= sz;
      else {
        k = sz + k;
        k %= sz;
      }
      rot[dep] += k;
      rot[dep] %= sz;
      if (op == 2) {
        long long incr = 2 * k;
        for (long long i = dep + 1; i <= 60; i++) {
          rot[i] += incr;
          rot[i] %= (1LL << i);
          incr *= 2;
        }
      }
      continue;
    }
    long long node;
    cin >> node;
    long long dep = log2l(node);
    long long pos_start = node - (1LL << dep);
    long long pos = (rot[dep] + pos_start) % (1LL << dep);
    vector<long long> out;
    while (dep > 0) {
      long long nxt = pos;
      long long incr = (1LL << dep) - rot[dep];
      long long new_pos = (pos + incr) % (1LL << dep);
      long long inp = (1LL << dep) + new_pos;
      out.push_back(inp);
      pos = nxt + (1LL << dep);
      pos /= 2;
      dep--;
    }
    out.push_back(1);
    for (auto x : out) cout << x << " ";
    cout << endl;
  }
}
