#include <bits/stdc++.h>
using namespace std;
long long shi[65] = {0}, son[65];
void init() {
  long long i;
  son[1] = 1;
  for (i = 2; i < 65; i++) {
    son[i] = son[i - 1] + son[i - 1];
  }
}
int H(long long X) {
  int cnt = 0;
  while (X) {
    cnt++;
    X >>= 1;
  }
  return cnt;
}
void find_root(int F, long long pos) {
  if (F == 0) return;
  long long A = pos / 2;
  pos -= shi[F];
  if (pos < 0)
    pos += son[F];
  else
    pos %= son[F];
  printf("%lld ", son[F] + pos);
  find_root(F - 1, A);
}
int main(void) {
  init();
  long long Q, T, X, F;
  scanf("%lld", &Q);
  while (Q--) {
    scanf("%lld", &T);
    if (T == 3) {
      scanf("%lld", &X);
      F = H(X);
      find_root(F, X - son[F] + shi[F]);
      printf("\n");
    } else if (T == 2) {
      scanf("%lld %lld", &X, &T);
      F = H(X);
      for (; F < 65; F++) {
        T = T % son[F] + son[F];
        shi[F] += T;
        if (shi[F] >= son[F]) shi[F] %= son[F];
        T *= 2;
      }
    } else {
      scanf("%lld %lld", &X, &T);
      F = H(X);
      T = T % son[F] + son[F];
      shi[F] += T;
      if (shi[F] < 0)
        shi[F] += son[F];
      else if (shi[F] >= son[F])
        shi[F] %= son[F];
    }
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 shi[65] = {0}, son[65];
void init() {
  long long i;
  son[1] = 1;
  for (i = 2; i < 65; i++) {
    son[i] = son[i - 1] + son[i - 1];
  }
}
int H(long long X) {
  int cnt = 0;
  while (X) {
    cnt++;
    X >>= 1;
  }
  return cnt;
}
void find_root(int F, long long pos) {
  if (F == 0) return;
  long long A = pos / 2;
  pos -= shi[F];
  if (pos < 0)
    pos += son[F];
  else
    pos %= son[F];
  printf("%lld ", son[F] + pos);
  find_root(F - 1, A);
}
int main(void) {
  init();
  long long Q, T, X, F;
  scanf("%lld", &Q);
  while (Q--) {
    scanf("%lld", &T);
    if (T == 3) {
      scanf("%lld", &X);
      F = H(X);
      find_root(F, X - son[F] + shi[F]);
      printf("\n");
    } else if (T == 2) {
      scanf("%lld %lld", &X, &T);
      F = H(X);
      for (; F < 65; F++) {
        T = T % son[F] + son[F];
        shi[F] += T;
        if (shi[F] >= son[F]) shi[F] %= son[F];
        T *= 2;
      }
    } else {
      scanf("%lld %lld", &X, &T);
      F = H(X);
      T = T % son[F] + son[F];
      shi[F] += T;
      if (shi[F] < 0)
        shi[F] += son[F];
      else if (shi[F] >= son[F])
        shi[F] %= son[F];
    }
  }
}

```