#include <bits/stdc++.h>
using namespace std;
long long decal[1000];
void get_path(long long x) {
  int lvl(0);
  while (x >= (1ll << (lvl + 1))) lvl++;
  bool beg(1);
  set<int> s;
  while (x) {
    assert(x < (1ll << (lvl + 1)));
    s.insert(x);
    if (!beg)
      x -= decal[lvl];
    else
      beg = 0;
    while (x < (1ll << lvl)) x += (1ll << lvl);
    while (x >= (1ll << (lvl + 1))) x -= (1ll << lvl);
    printf("%lld ", x);
    x += decal[lvl];
    while (x >= (1ll << (lvl + 1))) x -= (1ll << lvl);
    while (x < (1ll << lvl)) x += (1ll << lvl);
    x /= 2;
    lvl--;
    assert(s.find(x) == s.end());
  }
  printf("\n");
}
void update(int st, int dr, long long val) {
  assert(st <= dr);
  val = (val % (1ll << st) + (1ll << st)) % (1ll << st);
  for (int i(st); i <= dr; i++) {
    val %= (1ll << i);
    decal[i] += val;
    decal[i] %= (1ll << i);
    val *= 2;
  }
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  long long q;
  scanf("%lld", &q);
  while (q--) {
    long long t;
    scanf("%lld", &t);
    if (t == 1 || t == 2) {
      long long x, lvl(0), k;
      scanf("%lld%lld", &x, &k);
      while (x >= (1ll << (lvl + 1))) lvl++;
      update(lvl, (t == 2 ? 60 : lvl), k);
    } else {
      long long x;
      scanf("%lld", &x);
      get_path(x);
    }
  }
  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;
long long decal[1000];
void get_path(long long x) {
  int lvl(0);
  while (x >= (1ll << (lvl + 1))) lvl++;
  bool beg(1);
  set<int> s;
  while (x) {
    assert(x < (1ll << (lvl + 1)));
    s.insert(x);
    if (!beg)
      x -= decal[lvl];
    else
      beg = 0;
    while (x < (1ll << lvl)) x += (1ll << lvl);
    while (x >= (1ll << (lvl + 1))) x -= (1ll << lvl);
    printf("%lld ", x);
    x += decal[lvl];
    while (x >= (1ll << (lvl + 1))) x -= (1ll << lvl);
    while (x < (1ll << lvl)) x += (1ll << lvl);
    x /= 2;
    lvl--;
    assert(s.find(x) == s.end());
  }
  printf("\n");
}
void update(int st, int dr, long long val) {
  assert(st <= dr);
  val = (val % (1ll << st) + (1ll << st)) % (1ll << st);
  for (int i(st); i <= dr; i++) {
    val %= (1ll << i);
    decal[i] += val;
    decal[i] %= (1ll << i);
    val *= 2;
  }
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  long long q;
  scanf("%lld", &q);
  while (q--) {
    long long t;
    scanf("%lld", &t);
    if (t == 1 || t == 2) {
      long long x, lvl(0), k;
      scanf("%lld%lld", &x, &k);
      while (x >= (1ll << (lvl + 1))) lvl++;
      update(lvl, (t == 2 ? 60 : lvl), k);
    } else {
      long long x;
      scanf("%lld", &x);
      get_path(x);
    }
  }
  return 0;
}

```