#include <bits/stdc++.h>
using namespace std;
long long L[70];
void ShiftLevel(long long lvl, long long val) {
  long long St = (1LL << (lvl - 1));
  val = (val % St) + St;
  L[lvl] = (L[lvl] + val) % St;
}
long long Adjust(long long lvl, long long T) {
  long long St = (1LL << (lvl - 1));
  long long Ed = (1LL << (lvl)) - 1;
  if (T > Ed) {
    T = St + (T % Ed) - 1;
  }
  if (T < St) {
    T = Ed - (St - T) + 1;
  }
  return T;
}
int main() {
  long long q, t, k, lvl;
  scanf("%lld", &q);
  while (q--) {
    scanf("%lld", &t);
    if (t == 1) {
      scanf("%lld", &lvl), scanf("%lld", &k);
      lvl = log2(lvl) + 1;
      ShiftLevel(lvl, k);
    } else if (t == 2) {
      scanf("%lld", &lvl), scanf("%lld", &k);
      lvl = log2(lvl) + 1;
      while (lvl <= 60LL) {
        ShiftLevel(lvl++, k);
        k = (k * 2);
      }
    } else {
      scanf("%lld", &k);
      lvl = log2(k) + 1;
      long long RealNode = Adjust(lvl, k + L[lvl]);
      printf("%lld ", k);
      while (lvl != 1) {
        --lvl;
        RealNode /= 2;
        long long T = RealNode - L[lvl];
        printf("%lld ", Adjust(lvl, T));
      }
      printf("\n");
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long L[70];
void ShiftLevel(long long lvl, long long val) {
  long long St = (1LL << (lvl - 1));
  val = (val % St) + St;
  L[lvl] = (L[lvl] + val) % St;
}
long long Adjust(long long lvl, long long T) {
  long long St = (1LL << (lvl - 1));
  long long Ed = (1LL << (lvl)) - 1;
  if (T > Ed) {
    T = St + (T % Ed) - 1;
  }
  if (T < St) {
    T = Ed - (St - T) + 1;
  }
  return T;
}
int main() {
  long long q, t, k, lvl;
  scanf("%lld", &q);
  while (q--) {
    scanf("%lld", &t);
    if (t == 1) {
      scanf("%lld", &lvl), scanf("%lld", &k);
      lvl = log2(lvl) + 1;
      ShiftLevel(lvl, k);
    } else if (t == 2) {
      scanf("%lld", &lvl), scanf("%lld", &k);
      lvl = log2(lvl) + 1;
      while (lvl <= 60LL) {
        ShiftLevel(lvl++, k);
        k = (k * 2);
      }
    } else {
      scanf("%lld", &k);
      lvl = log2(k) + 1;
      long long RealNode = Adjust(lvl, k + L[lvl]);
      printf("%lld ", k);
      while (lvl != 1) {
        --lvl;
        RealNode /= 2;
        long long T = RealNode - L[lvl];
        printf("%lld ", Adjust(lvl, T));
      }
      printf("\n");
    }
  }
  return 0;
}

```