#include <bits/stdc++.h>
using namespace std;
const int MAXN = 70;
int Q;
int type;
long long x, k;
long long level[MAXN];
vector<long long> answer;
int getLevel(long long x) {
  int answer = 0;
  while (x != 0) {
    x /= 2;
    answer++;
  }
  return answer - 1;
}
long long fix(long long x, long long len) {
  if (x >= len) return x % len;
  if (x >= 0 && x < len) return x;
  return x % len + len;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int i, j;
  long long update = 0, current;
  cin >> Q;
  for (int q = 0; q < Q; q++) {
    cin >> type;
    if (type == 1) {
      cin >> x >> k;
      current = getLevel(x);
      level[current] = (level[current] + k) % (1LL << (current));
    } else if (type == 2) {
      cin >> x >> k;
      current = getLevel(x);
      for (i = current; i < 63; i++) {
        update = k;
        for (j = current; j < i; j++) {
          update = (update * 2) % (1LL << (i));
        }
        level[i] = (level[i] + update) % (1LL << i);
      }
    } else {
      cin >> x;
      current = getLevel(x);
      x = fix(x - ((1LL << (current)) - 1) + level[current] - 1,
              (1LL << (current))) +
          ((1LL << (current)) - 1) + 1;
      while (current != 0) {
        cout << fix(x - ((1LL << (current)) - 1) - level[current] - 1,
                    (1LL << (current))) +
                    ((1LL << (current)) - 1) + 1
             << " ";
        x /= 2;
        current--;
      }
      cout << "1" << '\n';
    }
  }
}

### 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 = 70;
int Q;
int type;
long long x, k;
long long level[MAXN];
vector<long long> answer;
int getLevel(long long x) {
  int answer = 0;
  while (x != 0) {
    x /= 2;
    answer++;
  }
  return answer - 1;
}
long long fix(long long x, long long len) {
  if (x >= len) return x % len;
  if (x >= 0 && x < len) return x;
  return x % len + len;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int i, j;
  long long update = 0, current;
  cin >> Q;
  for (int q = 0; q < Q; q++) {
    cin >> type;
    if (type == 1) {
      cin >> x >> k;
      current = getLevel(x);
      level[current] = (level[current] + k) % (1LL << (current));
    } else if (type == 2) {
      cin >> x >> k;
      current = getLevel(x);
      for (i = current; i < 63; i++) {
        update = k;
        for (j = current; j < i; j++) {
          update = (update * 2) % (1LL << (i));
        }
        level[i] = (level[i] + update) % (1LL << i);
      }
    } else {
      cin >> x;
      current = getLevel(x);
      x = fix(x - ((1LL << (current)) - 1) + level[current] - 1,
              (1LL << (current))) +
          ((1LL << (current)) - 1) + 1;
      while (current != 0) {
        cout << fix(x - ((1LL << (current)) - 1) - level[current] - 1,
                    (1LL << (current))) +
                    ((1LL << (current)) - 1) + 1
             << " ";
        x /= 2;
        current--;
      }
      cout << "1" << '\n';
    }
  }
}

```