#include <bits/stdc++.h>
using namespace std;
int q;
unsigned long long t, x, k;
long long pos[100];
int main() {
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%llu", &t);
    if (t < 3) {
      scanf("%llu%llu", &x, &k);
      int j;
      for (j = 62; j >= 0; j--) {
        if ((1uLL << j) & x) {
          k = ((k % (1uLL << j)) + (1uLL << j)) % (1uLL << j);
          pos[j] = (pos[j] + k) % (1uLL << j);
          break;
        }
      }
      if (t == 2)
        for (int p = j + 1; p < 63; ++p) {
          k = k * 2 % (1ull << p);
          pos[p] = (pos[p] + k) % (1uLL << p);
        }
    } else {
      scanf("%llu", &x);
      for (int j = 62; j >= 0; j--) {
        if ((1uLL << j) & x) {
          unsigned long long cur = (x + pos[j]);
          while (cur >= (1LL << (j + 1))) cur -= (1LL << j);
          for (int k = j; k >= 0; k--) {
            unsigned long long ans = (cur - pos[k]);
            if (ans < (1uLL << k)) ans += 1uLL << k;
            printf("%llu ", ans);
            cur /= 2;
          }
          printf("\n");
          break;
        }
      }
    }
  }
}

### Prompt

Your challenge is to write 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;
int q;
unsigned long long t, x, k;
long long pos[100];
int main() {
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%llu", &t);
    if (t < 3) {
      scanf("%llu%llu", &x, &k);
      int j;
      for (j = 62; j >= 0; j--) {
        if ((1uLL << j) & x) {
          k = ((k % (1uLL << j)) + (1uLL << j)) % (1uLL << j);
          pos[j] = (pos[j] + k) % (1uLL << j);
          break;
        }
      }
      if (t == 2)
        for (int p = j + 1; p < 63; ++p) {
          k = k * 2 % (1ull << p);
          pos[p] = (pos[p] + k) % (1uLL << p);
        }
    } else {
      scanf("%llu", &x);
      for (int j = 62; j >= 0; j--) {
        if ((1uLL << j) & x) {
          unsigned long long cur = (x + pos[j]);
          while (cur >= (1LL << (j + 1))) cur -= (1LL << j);
          for (int k = j; k >= 0; k--) {
            unsigned long long ans = (cur - pos[k]);
            if (ans < (1uLL << k)) ans += 1uLL << k;
            printf("%llu ", ans);
            cur /= 2;
          }
          printf("\n");
          break;
        }
      }
    }
  }
}

```