#include <bits/stdc++.h>
using namespace std;
const int bit = 62;
int m;
long long shift[66];
int main() {
  scanf("%d", &m);
  int i, j, s, p, q, flag;
  long long t, x, k, y;
  int cnt;
  while (m--) {
    scanf("%d", &flag);
    if (flag <= 2) {
      scanf("%lld%lld", &x, &k);
      y = x;
      cnt = 0;
      while (y > 1) {
        y >>= 1;
        cnt++;
      }
      k %= (1LL << cnt);
      shift[cnt] = (shift[cnt] + k) % (1LL << cnt);
      if (shift[cnt] < 0) shift[cnt] += (1LL << cnt);
      if (flag == 2) {
        for (i = cnt + 1; i <= bit; i++) {
          k *= 2;
          shift[i] = (shift[i] + k) % (1LL << i);
          if (shift[cnt] < 0) shift[cnt] += (1LL << i);
        }
      }
    } else {
      scanf("%lld", &x);
      y = x;
      cnt = 0;
      while (y > 1) {
        y >>= 1;
        cnt++;
      }
      long long id = (x + shift[cnt]) % (1LL << cnt);
      id += (1LL << cnt);
      while (id) {
        printf("%lld", x);
        if (x == 1)
          break;
        else
          putchar(' ');
        id /= 2;
        cnt--;
        x = (id - shift[cnt]) % (1LL << cnt);
        x += (1LL << cnt);
      }
      printf("\n");
    }
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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 bit = 62;
int m;
long long shift[66];
int main() {
  scanf("%d", &m);
  int i, j, s, p, q, flag;
  long long t, x, k, y;
  int cnt;
  while (m--) {
    scanf("%d", &flag);
    if (flag <= 2) {
      scanf("%lld%lld", &x, &k);
      y = x;
      cnt = 0;
      while (y > 1) {
        y >>= 1;
        cnt++;
      }
      k %= (1LL << cnt);
      shift[cnt] = (shift[cnt] + k) % (1LL << cnt);
      if (shift[cnt] < 0) shift[cnt] += (1LL << cnt);
      if (flag == 2) {
        for (i = cnt + 1; i <= bit; i++) {
          k *= 2;
          shift[i] = (shift[i] + k) % (1LL << i);
          if (shift[cnt] < 0) shift[cnt] += (1LL << i);
        }
      }
    } else {
      scanf("%lld", &x);
      y = x;
      cnt = 0;
      while (y > 1) {
        y >>= 1;
        cnt++;
      }
      long long id = (x + shift[cnt]) % (1LL << cnt);
      id += (1LL << cnt);
      while (id) {
        printf("%lld", x);
        if (x == 1)
          break;
        else
          putchar(' ');
        id /= 2;
        cnt--;
        x = (id - shift[cnt]) % (1LL << cnt);
        x += (1LL << cnt);
      }
      printf("\n");
    }
  }
  return 0;
}

```