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() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int Q;
  cin >> Q;
  long long S[61];
  fill(S, S + 61, 0);
  while (Q--) {
    int T;
    long long X, K;
    cin >> T;
    cin >> X;
    int L = 64 - __builtin_clzll(X);
    if (T != 3) {
      cin >> K;
      int eL = 60;
      if (T == 1) eL = L;
      for (int i = 0; L + i <= eL; i++) {
        S[L + i] += (K << i);
        S[L + i] &= ((1LL << (L + i - 1)) - 1);
      }
    } else {
      long long sh = 1;
      sh <<= L - 1;
      long long P = X - sh;
      long long Ph = (((P + S[L]) & (sh - 1)) + sh) & (sh - 1);
      long long V = sh + Ph;
      V >>= 1;
      sh >>= 1;
      L--;
      cout << X;
      while (V) {
        P = V - sh;
        Ph = (((P - S[L]) & (sh - 1)) + sh) & (sh - 1);
        cout << ' ' << sh + Ph;
        L--;
        V >>= 1;
        sh >>= 1;
      }
      cout << '\n';
    }
  }
  return 0;
}