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 arr[65];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int q, i, j, temp, pos, cnt, x, k, t, p, m;
  long long int MOD = (long long int)pow(2, 60);
  cin >> q;
  for (i = 0; i < q; i++) {
    cin >> t;
    if (t == 1) {
      cin >> x >> k;
      temp = x;
      cnt = 0;
      while (temp != 0) {
        temp /= 2;
        cnt++;
      }
      m = pow(2, cnt - 1);
      arr[cnt] = (arr[cnt] + k + MOD + m) % m;
    } else if (t == 2) {
      cin >> x >> k;
      temp = x;
      cnt = 0;
      while (temp != 0) {
        temp /= 2;
        cnt++;
      }
      m = pow(2, cnt - 1);
      p = k;
      for (j = cnt; j <= 60; j++) {
        arr[j] = (arr[j] + p + MOD + m) % m;
        m = (m * 2) % MOD;
        p = (p * 2) % MOD;
      }
    } else {
      cin >> x;
      temp = x;
      cnt = 0;
      while (temp != 0) {
        temp /= 2;
        cnt++;
      }
      m = pow(2, cnt - 1);
      pos = ((x - m) + (arr[cnt]) % m) % m;
      while ((m + pos) != 0) {
        cout << m + (pos - arr[cnt] + m) % m << " ";
        pos = (m + pos) / 2 - m / 2;
        m = m / 2;
        cnt--;
      }
      cout << endl;
    }
  }
}