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 high(long long x) {
  long long sum = 0;
  while (x) {
    sum++;
    x /= 2;
  }
  return sum;
}
long long d1[66], d2[66];
long long fa(long long x) {
  if (x == 1) return 0;
  long long h = high(x);
  long long f = x + (d1[h] + d2[h]) - (1ll << (h - 1));
  f = ((f % (1ll << (h - 1))) + (1ll << (h - 1))) % (1ll << (h - 1));
  f += (1ll << (h - 1));
  f = f / 2;
  h--;
  f = f - (d1[h]) - (1ll << (h - 1));
  f = ((f % (1ll << (h - 1))) + (1ll << (h - 1))) % (1ll << (h - 1));
  f += (1ll << (h - 1));
  return f;
}
int main() {
  long long q;
  scanf("%I64d", &q);
  while (q--) {
    long long op;
    scanf("%I64d", &op);
    if (op == 1) {
      long long x, k;
      scanf("%I64d%I64d", &x, &k);
      long long h = high(x);
      d1[h] = (d1[h] + k) % (1ll << (h - 1));
    } else if (op == 2) {
      long long x, k;
      scanf("%I64d%I64d", &x, &k);
      long long h = high(x);
      d2[h] = (d2[h] + k) % (1ll << (h - 1));
    } else if (op == 3) {
      long long x;
      scanf("%I64d", &x);
      while (x > 1) {
        printf("%I64d ", x);
        x = fa(x);
      }
      printf("%I64d\n", x);
    }
  }
  return 0;
}