#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
long long a[70], b[70];
int get(long long x) {
  int m = -1;
  while (x) {
    m++;
    x >>= 1;
  }
  return m;
}
int main() {
  int q;
  scanf("%d", &q);
  for (int i = 0; i < 62; i++) {
    b[i] = 1ll << i;
  }
  while (q--) {
    long long x, k;
    int f;
    scanf("%d", &f);
    if (f == 1) {
      scanf("%I64d%I64d", &x, &k);
      int pos = get(x);
      k = (k % b[pos] + b[pos]) % b[pos];
      a[pos] = (a[pos] + k + b[pos]) % b[pos];
    } else if (f == 2) {
      scanf("%I64d%I64d", &x, &k);
      int pos = get(x);
      k = (k % b[pos] + b[pos]) % b[pos];
      for (int i = pos; i < 62; i++) {
        a[i] = (a[i] + k + b[i]) % b[i];
        k = k * 2;
      }
    } else {
      scanf("%I64d", &x);
      printf("%I64d", x);
      long long pos = get(x);
      int te = pos;
      pos = (x - b[pos] + a[pos]) % b[pos];
      for (int i = te - 1; i >= 0; i--) {
        pos = pos / 2;
        printf(" %I64d", (pos - a[i] + b[i]) % b[i] + b[i]);
      }
      puts("");
    }
  }
}

### 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;
const int maxn = 1e6 + 5;
long long a[70], b[70];
int get(long long x) {
  int m = -1;
  while (x) {
    m++;
    x >>= 1;
  }
  return m;
}
int main() {
  int q;
  scanf("%d", &q);
  for (int i = 0; i < 62; i++) {
    b[i] = 1ll << i;
  }
  while (q--) {
    long long x, k;
    int f;
    scanf("%d", &f);
    if (f == 1) {
      scanf("%I64d%I64d", &x, &k);
      int pos = get(x);
      k = (k % b[pos] + b[pos]) % b[pos];
      a[pos] = (a[pos] + k + b[pos]) % b[pos];
    } else if (f == 2) {
      scanf("%I64d%I64d", &x, &k);
      int pos = get(x);
      k = (k % b[pos] + b[pos]) % b[pos];
      for (int i = pos; i < 62; i++) {
        a[i] = (a[i] + k + b[i]) % b[i];
        k = k * 2;
      }
    } else {
      scanf("%I64d", &x);
      printf("%I64d", x);
      long long pos = get(x);
      int te = pos;
      pos = (x - b[pos] + a[pos]) % b[pos];
      for (int i = te - 1; i >= 0; i--) {
        pos = pos / 2;
        printf(" %I64d", (pos - a[i] + b[i]) % b[i] + b[i]);
      }
      puts("");
    }
  }
}

```