#include <bits/stdc++.h>
using namespace std;
int q;
long long t, x, k;
long long pow2[63], lmost[63];
void init() {
  long long res = 1;
  for (int i = 0; i < 62; i++) lmost[i] = pow2[i] = res, res <<= 1;
}
void q1(long long x, long long k) {
  int level = 0;
  while (pow2[level + 1] <= x) {
    level++;
  }
  lmost[level] = (-k + lmost[level] + pow2[level]) % pow2[level] + pow2[level];
}
void q2(long long x, long long k) {
  int level = 0;
  while (pow2[level + 1] <= x) {
    level++;
  }
  while (level < 62)
    lmost[level] =
        (-k + lmost[level] + pow2[level]) % pow2[level] + pow2[level],
    k = (k + pow2[level]) % pow2[level] * 2, level++;
}
void q3(long long x) {
  int level = 0;
  long long curloc = 0;
  while (pow2[level + 1] <= x) {
    level++;
  }
  curloc = (x - lmost[level] + pow2[level]) % pow2[level];
  while (level >= 0) {
    long long now = (curloc + lmost[level]) % pow2[level] + pow2[level];
    printf("%I64d ", now);
    level--;
    curloc /= 2;
  }
  printf("\n");
}
int main() {
  init();
  scanf("%d", &q);
  while (q--) {
    scanf("%I64d", &t);
    if (t == 1) {
      scanf("%I64d%I64d", &x, &k);
      q1(x, k);
    } else if (t == 2) {
      scanf("%I64d%I64d", &x, &k);
      q2(x, k);
    } else {
      scanf("%I64d", &x);
      q3(x);
    }
  }
}

### 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;
long long t, x, k;
long long pow2[63], lmost[63];
void init() {
  long long res = 1;
  for (int i = 0; i < 62; i++) lmost[i] = pow2[i] = res, res <<= 1;
}
void q1(long long x, long long k) {
  int level = 0;
  while (pow2[level + 1] <= x) {
    level++;
  }
  lmost[level] = (-k + lmost[level] + pow2[level]) % pow2[level] + pow2[level];
}
void q2(long long x, long long k) {
  int level = 0;
  while (pow2[level + 1] <= x) {
    level++;
  }
  while (level < 62)
    lmost[level] =
        (-k + lmost[level] + pow2[level]) % pow2[level] + pow2[level],
    k = (k + pow2[level]) % pow2[level] * 2, level++;
}
void q3(long long x) {
  int level = 0;
  long long curloc = 0;
  while (pow2[level + 1] <= x) {
    level++;
  }
  curloc = (x - lmost[level] + pow2[level]) % pow2[level];
  while (level >= 0) {
    long long now = (curloc + lmost[level]) % pow2[level] + pow2[level];
    printf("%I64d ", now);
    level--;
    curloc /= 2;
  }
  printf("\n");
}
int main() {
  init();
  scanf("%d", &q);
  while (q--) {
    scanf("%I64d", &t);
    if (t == 1) {
      scanf("%I64d%I64d", &x, &k);
      q1(x, k);
    } else if (t == 2) {
      scanf("%I64d%I64d", &x, &k);
      q2(x, k);
    } else {
      scanf("%I64d", &x);
      q3(x);
    }
  }
}

```