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 par[101];
vector<int> vec;
int main() {
  int q;
  scanf("%d", &q);
  while (q--) {
    int t;
    long long x, k;
    scanf("%d %lld", &t, &x);
    if (t == 1) {
      scanf("%lld", &k);
      int d = 0;
      long long node = 1;
      while (node * 2 <= x) {
        node += node;
        d++;
      }
      par[d] = (par[d] + k) % node;
      par[d] = (par[d] + node) % node;
      par[d + 1] = (par[d + 1] - 2 * k) % (2 * node);
      par[d + 1] = (par[d + 1] + (2 * node)) % (2 * node);
    } else if (t == 2) {
      scanf("%lld", &k);
      int d = 0;
      long long node = 1;
      while (node * 2 <= x) {
        node += node;
        d++;
      }
      par[d] = (par[d] + k) % node;
      par[d] = (par[d] + node) % node;
    } else {
      vec.clear();
      int d = 0;
      long long node = 1;
      while (node * 2 <= x) {
        node += node;
        d++;
      }
      while (x != 1) {
        printf("%lld ", x);
        if (x + par[d] >= node * 2) {
          x = x + par[d] - node;
          x /= 2;
        } else {
          x = x + par[d];
          x /= 2;
        }
        d--;
        node /= 2;
      }
      printf("1\n");
    }
  }
}