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 level[70][3], shift[70], par[70];
void print(long long int x, long long int i) {
  long long int tmp = (x - shift[i]);
  if (tmp < level[i][0])
    tmp = (level[i][1] + 1 - level[i][0] + tmp);
  else if (tmp > level[i][1])
    tmp = (level[i][0] - 1 - level[i][1] + tmp);
  printf("%lld ", tmp);
  if (x == 1) return;
  x += par[i] % (level[i][1] - level[i][0] + 1);
  if (x < level[i][0])
    x = (level[i][1] + 1 - level[i][0] + x);
  else if (x > level[i][1])
    x = (level[i][0] - 1 - level[i][1] + x);
  print(x / 2, i - 1);
}
int main() {
  long long int x, k, i;
  int q, t;
  for (i = 0; i < 60; i++) {
    level[i][0] = (1ll << i);
    level[i][1] = (1ll << (i + 1)) - 1;
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d %lld", &t, &x);
    for (i = 0;; i++)
      if (level[i][0] <= x && x <= level[i][1]) break;
    if (t == 3) {
      x += shift[i];
      if (x < level[i][0])
        x = (level[i][1] + 1 - level[i][0] + x);
      else if (x > level[i][1])
        x = (level[i][0] - 1 - level[i][1] + x);
      print(x, i);
      printf("\n");
    } else {
      scanf("%lld", &k);
      if (t == 1) {
        shift[i] = (k + shift[i]) % (level[i][1] - level[i][0] + 1);
      } else
        par[i] = (k + par[i]) % (level[i][1] - level[i][0] + 1);
    }
  }
  return 0;
}