#include <bits/stdc++.h>
using namespace std;
const int maxk = 64;
const int maxn = 1e5 + 17;
int Q;
long long size[maxk];
long long offset[maxk];
inline void shift(int level, long long &x) {
  if (x < 0) x = size[level] - abs(x) % size[level];
  offset[level] = (offset[level] + (x %= size[level])) % size[level];
}
inline void getValue(int level, long long where) {
  printf("%lld ",
         (size[level] - offset[level] + where) % size[level] + size[level]);
  if (!level) printf("\n");
}
int main() {
  for (int i = 0; i < maxk; i++) size[i] = 1LL << i;
  scanf("%d", &Q);
  while (Q--) {
    int t;
    long long x, k;
    scanf("%d %lld", &t, &x);
    if (t == 1 || t == 2) scanf("%lld", &k);
    int level = 8 * sizeof(long long) - __builtin_clzll(x) - 1;
    if (t == 1) shift(level, k);
    if (t == 2)
      for (int i = level; i < maxk; i++) shift(i, k), k <<= 1;
    if (t == 3) {
      long long idx = (x - size[level] + offset[level]) % size[level];
      while (level >= 0) {
        getValue(level, idx);
        level--;
        idx >>= 1;
      }
    }
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following 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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxk = 64;
const int maxn = 1e5 + 17;
int Q;
long long size[maxk];
long long offset[maxk];
inline void shift(int level, long long &x) {
  if (x < 0) x = size[level] - abs(x) % size[level];
  offset[level] = (offset[level] + (x %= size[level])) % size[level];
}
inline void getValue(int level, long long where) {
  printf("%lld ",
         (size[level] - offset[level] + where) % size[level] + size[level]);
  if (!level) printf("\n");
}
int main() {
  for (int i = 0; i < maxk; i++) size[i] = 1LL << i;
  scanf("%d", &Q);
  while (Q--) {
    int t;
    long long x, k;
    scanf("%d %lld", &t, &x);
    if (t == 1 || t == 2) scanf("%lld", &k);
    int level = 8 * sizeof(long long) - __builtin_clzll(x) - 1;
    if (t == 1) shift(level, k);
    if (t == 2)
      for (int i = level; i < maxk; i++) shift(i, k), k <<= 1;
    if (t == 3) {
      long long idx = (x - size[level] + offset[level]) % size[level];
      while (level >= 0) {
        getValue(level, idx);
        level--;
        idx >>= 1;
      }
    }
  }
  return 0;
}

```