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 q, t, x, l, k, n, temp, v, rs[100], pos;
int main() {
  cin >> q;
  while (q--) {
    cin >> t >> x;
    l = log2(x) + 1;
    if (t == 1) {
      cin >> k;
      n = 1LL << (l - 1);
      temp = (k % n + n) % n;
      rs[l] = (rs[l] + temp) % n;
    } else if (t == 2) {
      cin >> k;
      n = 1LL << (l - 1);
      temp = (k % n + n) % n;
      for (int i = l; i <= 61; i++) {
        n = 1LL << (i - 1);
        temp %= n;
        rs[i] = (rs[i] + temp) % n;
        temp *= 2;
      }
    } else {
      n = 1LL << (l - 1);
      pos = x - n + 1;
      pos = (pos + rs[l]) % n;
      if (pos == 0) pos = n;
      v = n + pos - 1;
      v /= 2;
      l--;
      cout << x << " ";
      while (v) {
        n = (1LL << (l - 1));
        pos = v - n + 1;
        pos = ((pos - rs[l]) % n + n) % n;
        if (pos == 0) pos = n;
        cout << n + pos - 1 << " ";
        v /= 2;
        l--;
      }
      cout << endl;
    }
  }
  return 0;
}