#include <bits/stdc++.h>
using namespace std;
map<long long, long long> Q1, Q2;
long long newPos(long long pos, long long shift, long long lvlPow2) {
  lvlPow2 /= 2;
  if (lvlPow2 == 1) return 1;
  shift %= lvlPow2;
  if (0 > shift) shift += lvlPow2;
  if (pos - lvlPow2 - shift >= 0) {
    return (pos - shift);
  } else {
    return pos + lvlPow2 - shift;
  }
}
long long orginPos(long long pos, long long shift, long long lvlPow2) {
  lvlPow2 /= 2;
  if (lvlPow2 == 1) return 1;
  shift %= lvlPow2;
  return (pos + shift) % lvlPow2 + lvlPow2;
}
void Q3(long long node) {
  printf("%lld ", node);
  long long lvl, lvlPow2, shift;
  lvl = log2(node) + 1, lvlPow2 = 1ll << lvl;
  long long pos = orginPos(node, Q1[lvl] + Q2[lvl], lvlPow2) / 2;
  while (pos) {
    lvl--, lvlPow2 /= 2, shift = Q1[lvl];
    printf("%lld ", newPos(pos, shift, lvlPow2));
    pos = orginPos(pos, Q2[lvl], lvlPow2) / 2;
  }
  puts("");
}
int main() {
  int n, QType;
  long long a, b;
  cin >> n;
  while (n--) {
    scanf("%d", &QType);
    if (QType == 1) {
      scanf("%lld%lld", &a, &b);
      if (a == 1) continue;
      a = log2(a) + 1;
      Q1[a] += b;
    } else if (QType == 2) {
      scanf("%lld%lld", &a, &b);
      a = log2(a) + 1;
      Q2[a] += b;
    } else {
      scanf("%lld", &a);
      Q3(a);
    }
  }
  return 0;
}

### 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;
map<long long, long long> Q1, Q2;
long long newPos(long long pos, long long shift, long long lvlPow2) {
  lvlPow2 /= 2;
  if (lvlPow2 == 1) return 1;
  shift %= lvlPow2;
  if (0 > shift) shift += lvlPow2;
  if (pos - lvlPow2 - shift >= 0) {
    return (pos - shift);
  } else {
    return pos + lvlPow2 - shift;
  }
}
long long orginPos(long long pos, long long shift, long long lvlPow2) {
  lvlPow2 /= 2;
  if (lvlPow2 == 1) return 1;
  shift %= lvlPow2;
  return (pos + shift) % lvlPow2 + lvlPow2;
}
void Q3(long long node) {
  printf("%lld ", node);
  long long lvl, lvlPow2, shift;
  lvl = log2(node) + 1, lvlPow2 = 1ll << lvl;
  long long pos = orginPos(node, Q1[lvl] + Q2[lvl], lvlPow2) / 2;
  while (pos) {
    lvl--, lvlPow2 /= 2, shift = Q1[lvl];
    printf("%lld ", newPos(pos, shift, lvlPow2));
    pos = orginPos(pos, Q2[lvl], lvlPow2) / 2;
  }
  puts("");
}
int main() {
  int n, QType;
  long long a, b;
  cin >> n;
  while (n--) {
    scanf("%d", &QType);
    if (QType == 1) {
      scanf("%lld%lld", &a, &b);
      if (a == 1) continue;
      a = log2(a) + 1;
      Q1[a] += b;
    } else if (QType == 2) {
      scanf("%lld%lld", &a, &b);
      a = log2(a) + 1;
      Q2[a] += b;
    } else {
      scanf("%lld", &a);
      Q3(a);
    }
  }
  return 0;
}

```