#include <bits/stdc++.h>
using namespace std;
long long A[69], nodes[69], maxVal[69];
int getHt(long long x) {
  int ans = 0;
  while (x) x /= 2, ans++;
  return ans - 1;
}
long long getNode(long long x, long long s, int h) {
  x += s;
  if (x <= maxVal[h]) return x;
  x = maxVal[h - 1] + (x - maxVal[h]);
  return x;
}
void solve(long long x, int h) {
  cout << getNode(x, nodes[h] - A[h], h) << " ";
  if (x == 1) return;
  solve(x / 2, h - 1);
}
int main() {
  cin.tie(NULL), cout.tie(NULL);
  int Q;
  cin >> Q;
  nodes[0] = maxVal[0] = 1;
  for (int i = 1; i <= 60; i++) {
    nodes[i] = nodes[i - 1] * 2;
    maxVal[i] = maxVal[i - 1] + nodes[i];
  }
  while (Q--) {
    int type;
    cin >> type;
    if (type == 1) {
      long long x, k;
      cin >> x >> k;
      int h = getHt(x);
      k = (k % nodes[h] + nodes[h]) % nodes[h];
      A[h] = (A[h] + k) % nodes[h];
    } else if (type == 2) {
      long long x, k;
      cin >> x >> k;
      for (int h = getHt(x); h <= 60; h++, k = k * 2) {
        k = (k % nodes[h] + nodes[h]) % nodes[h];
        A[h] = (A[h] + k) % nodes[h];
      }
    } else {
      long long x;
      cin >> x;
      int h = getHt(x);
      long long y = getNode(x, A[h], h);
      solve(y, h);
      cout << "\n";
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
long long A[69], nodes[69], maxVal[69];
int getHt(long long x) {
  int ans = 0;
  while (x) x /= 2, ans++;
  return ans - 1;
}
long long getNode(long long x, long long s, int h) {
  x += s;
  if (x <= maxVal[h]) return x;
  x = maxVal[h - 1] + (x - maxVal[h]);
  return x;
}
void solve(long long x, int h) {
  cout << getNode(x, nodes[h] - A[h], h) << " ";
  if (x == 1) return;
  solve(x / 2, h - 1);
}
int main() {
  cin.tie(NULL), cout.tie(NULL);
  int Q;
  cin >> Q;
  nodes[0] = maxVal[0] = 1;
  for (int i = 1; i <= 60; i++) {
    nodes[i] = nodes[i - 1] * 2;
    maxVal[i] = maxVal[i - 1] + nodes[i];
  }
  while (Q--) {
    int type;
    cin >> type;
    if (type == 1) {
      long long x, k;
      cin >> x >> k;
      int h = getHt(x);
      k = (k % nodes[h] + nodes[h]) % nodes[h];
      A[h] = (A[h] + k) % nodes[h];
    } else if (type == 2) {
      long long x, k;
      cin >> x >> k;
      for (int h = getHt(x); h <= 60; h++, k = k * 2) {
        k = (k % nodes[h] + nodes[h]) % nodes[h];
        A[h] = (A[h] + k) % nodes[h];
      }
    } else {
      long long x;
      cin >> x;
      int h = getHt(x);
      long long y = getNode(x, A[h], h);
      solve(y, h);
      cout << "\n";
    }
  }
  return 0;
}

```