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 modexpo(long long int x, long long int y) {
  if (y == 0) return 1;
  if (y % 2) {
    long long int viky = modexpo(x, y / 2);
    return (((x * viky) % 1000000007) * viky) % 1000000007;
  } else {
    long long int viky = modexpo(x, y / 2);
    return (viky * viky) % 1000000007;
  }
}
long long int level[66];
long long int get_level(long long int x) {
  long long int c = 0;
  while (x) {
    c++;
    x = x / 2;
  }
  return c;
}
long long int level_update[66];
int main() {
  long long int t, i, j, k, s, l, r, n, x;
  level[0] = 1;
  level[1] = 1;
  for (i = 2; i <= 60; i++) {
    level[i] = 2 * level[i - 1];
  }
  scanf("%lld", &n);
  while (n--) {
    scanf("%lld", &l);
    if (l == 1) {
      scanf("%lld%lld", &x, &k);
      level_update[get_level(x)] += k;
    } else if (l == 2) {
      scanf("%lld%lld", &x, &k);
      long long int p = 1;
      for (i = get_level(x); i <= 60; i++) {
        level_update[i] += p * k;
        p = p * 2;
      }
    } else {
      scanf("%lld", &x);
      long long int xp = get_level(x);
      long long int o_index = x - level[xp];
      long long int curr_index =
          (o_index + (level_update[xp] % level[xp]) + level[xp]) % level[xp];
      printf("%lld ", x);
      while (--xp) {
        long long int par_curr = curr_index / 2;
        long long int lp = par_curr;
        curr_index =
            (par_curr - (level_update[xp] % level[xp]) + level[xp]) % level[xp];
        printf("%lld ", (curr_index + level[xp]));
        curr_index = lp;
      }
      printf("\n");
    }
  }
  return 0;
}