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;
vector<long long> pows(63);
int get_level(long long x) {
  auto it = upper_bound(pows.begin(), pows.end(), x);
  it--;
  return it - pows.begin();
}
int main() {
  pows[0] = 1;
  for (int i = 0; i < 63; i++) {
    if (i) {
      pows[i] = pows[i - 1] << 1;
    }
  }
  int q;
  scanf("%d", &q);
  vector<long long> val_shift(63), node_shift(63);
  for (int i = 0; i < q; i++) {
    int t;
    long long x;
    scanf("%d%llu", &t, &x);
    if (t == 1) {
      long long k;
      scanf("%llu", &k);
      int level = get_level(x);
      long long level_size = pows[level];
      k %= level_size;
      val_shift[level] = (val_shift[level] - k + level_size) % level_size;
    } else if (t == 2) {
      long long k;
      scanf("%llu", &k);
      int level = get_level(x);
      long long level_size = pows[level];
      k %= level_size;
      node_shift[level] = (node_shift[level] - k + level_size) % level_size;
    } else {
      long long cur = x;
      int start_level = get_level(x);
      long long level_size = pows[start_level];
      for (int level = start_level; level >= 0; level--) {
        printf("%llu ", cur);
        if (level == 0) {
          break;
        }
        cur -= val_shift[level];
        if (cur < level_size) {
          cur += level_size;
        }
        cur -= node_shift[level];
        if (cur < level_size) {
          cur += level_size;
        }
        cur >>= 1LL;
        int next_level = level - 1;
        long long next_level_size = level_size >> 1LL;
        cur += val_shift[next_level];
        if (cur >= level_size) {
          cur -= next_level_size;
        }
        level_size = next_level_size;
      }
      printf("\n");
    }
  }
}