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 higbit(long long x) {
  long long a, b, c, d;
  a = 0;
  b = x;
  while (x > 0) {
    x /= 2;
    a++;
  }
  return a - 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long a, b, c, d, e, i, j, k;
  long long q, t, x;
  cin >> q;
  long long sh1[100], sh2[100];
  for (i = 0; i < 100; i++) sh1[i] = 0, sh2[i] = 0;
  while (q--) {
    cin >> t;
    if (t == 1) {
      cin >> x >> k;
      a = higbit(x);
      b = 1LL << a;
      if (k < 0) {
        k = b - ((-k) % b);
      };
      k %= b;
      sh1[a] = (sh1[a] + k) % b;
    } else if (t == 2) {
      cin >> x >> k;
      a = higbit(x);
      b = 1LL << a;
      if (k < 0) {
        k = b - ((-1 * k) % b);
      };
      k %= b;
      sh2[a] = (sh2[a] + k) % b;
    } else {
      cin >> x;
      a = higbit(x);
      b = 1LL << a;
      while (x > 1) {
        cout << x << " ";
        c = (sh1[a] + sh2[a]) % b;
        d = x - b;
        d = (d + c) % b;
        d += b;
        d /= 2;
        b /= 2;
        a--;
        c = sh1[a];
        e = d - b;
        e = (e - c + b) % b;
        e += b;
        x = e;
      }
      cout << 1 << "\n";
    }
  }
  return 0;
}