#include <bits/stdc++.h>
using namespace std;
int Q;
long long bit[61];
long long tag[61];
int get_dep(long long x) {
  int h = 1;
  while (x != 1) x >>= 1, h++;
  return h;
}
int main() {
  bit[0] = 1;
  for (int i = 1; i <= 60; i++) bit[i] = bit[i - 1] << 1;
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) {
    int opt;
    long long x, k;
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%lld%lld", &x, &k);
      int dep = get_dep(x);
      tag[dep] = ((tag[dep] + k) % bit[dep - 1] + bit[dep - 1]) % bit[dep - 1];
    }
    if (opt == 2) {
      scanf("%lld%lld", &x, &k);
      int dep = get_dep(x);
      for (int j = dep; j <= 60; j++) {
        tag[j] = ((tag[j] + k) % bit[j - 1] + bit[j - 1]) % bit[j - 1];
        k = (k << 1) % bit[60];
      }
    }
    if (opt == 3) {
      scanf("%lld", &x);
      int dep = get_dep(x);
      long long u = x - bit[dep - 1];
      u = (u + tag[dep]) % bit[dep - 1];
      x = u + bit[dep - 1];
      for (int j = dep; j >= 1; j--) {
        long long v = x - bit[j - 1];
        v = ((v - tag[j]) % bit[j - 1] + bit[j - 1]) % bit[j - 1];
        v = v + bit[j - 1];
        printf("%lld ", v);
        x >>= 1;
      }
      puts("");
    }
  }
  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;
int Q;
long long bit[61];
long long tag[61];
int get_dep(long long x) {
  int h = 1;
  while (x != 1) x >>= 1, h++;
  return h;
}
int main() {
  bit[0] = 1;
  for (int i = 1; i <= 60; i++) bit[i] = bit[i - 1] << 1;
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) {
    int opt;
    long long x, k;
    scanf("%d", &opt);
    if (opt == 1) {
      scanf("%lld%lld", &x, &k);
      int dep = get_dep(x);
      tag[dep] = ((tag[dep] + k) % bit[dep - 1] + bit[dep - 1]) % bit[dep - 1];
    }
    if (opt == 2) {
      scanf("%lld%lld", &x, &k);
      int dep = get_dep(x);
      for (int j = dep; j <= 60; j++) {
        tag[j] = ((tag[j] + k) % bit[j - 1] + bit[j - 1]) % bit[j - 1];
        k = (k << 1) % bit[60];
      }
    }
    if (opt == 3) {
      scanf("%lld", &x);
      int dep = get_dep(x);
      long long u = x - bit[dep - 1];
      u = (u + tag[dep]) % bit[dep - 1];
      x = u + bit[dep - 1];
      for (int j = dep; j >= 1; j--) {
        long long v = x - bit[j - 1];
        v = ((v - tag[j]) % bit[j - 1] + bit[j - 1]) % bit[j - 1];
        v = v + bit[j - 1];
        printf("%lld ", v);
        x >>= 1;
      }
      puts("");
    }
  }
  return 0;
}

```