#include <bits/stdc++.h>
using namespace std;
int level(long long node) {
  if (node >> 32) return 31 - __builtin_clz(node >> 32) + 32;
  return 31 - __builtin_clz((int)node);
}
pair<long long, long long> bounds(int level) {
  return {(1LL << level), (1LL << level + 1) - 1};
}
long long posmod(long long x, long long m) {
  if (x > 0) return x;
  return (x + m * (-x / m + 1)) % m;
}
int main() {
  ios::sync_with_stdio(0);
  int Q;
  cin >> Q;
  vector<long long> nshift(100, 0);
  int type;
  long long node, shift;
  while (Q--) {
    cin >> type >> node;
    if (type < 3) {
      cin >> shift;
      int lev = level(node);
      auto bound = bounds(lev);
      nshift[lev] = (nshift[lev] + shift) % (bound.second - bound.first + 1);
      if (type == 2 && shift != 0) {
        long long cshift = 2 * shift;
        int i = 1;
        for (; (1ll << lev + i) < 1.2e18; i++, cshift *= 2) {
          nshift[lev + i] = (nshift[lev + i] + posmod(cshift, 1ll << lev + i)) %
                            (1ll << lev + i);
        }
      }
    } else {
      cout << node << ' ';
      auto bound = bounds(level(node));
      long long nlevels = bound.second - bound.first + 1;
      long long zeroorig = node - nlevels + nshift[level(node)];
      if (zeroorig < 0) zeroorig += 2 * nlevels;
      long long orig = zeroorig % nlevels + bound.first;
      while (orig / 2) {
        orig /= 2;
        auto bound = bounds(level(orig));
        long long nlevels = bound.second - bound.first + 1;
        cout << (orig - nlevels - nshift[level(orig)] + 2 * nlevels) % nlevels +
                    bound.first
             << ' ';
      }
      cout << '\n';
    }
  }
}

### Prompt

Generate 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 level(long long node) {
  if (node >> 32) return 31 - __builtin_clz(node >> 32) + 32;
  return 31 - __builtin_clz((int)node);
}
pair<long long, long long> bounds(int level) {
  return {(1LL << level), (1LL << level + 1) - 1};
}
long long posmod(long long x, long long m) {
  if (x > 0) return x;
  return (x + m * (-x / m + 1)) % m;
}
int main() {
  ios::sync_with_stdio(0);
  int Q;
  cin >> Q;
  vector<long long> nshift(100, 0);
  int type;
  long long node, shift;
  while (Q--) {
    cin >> type >> node;
    if (type < 3) {
      cin >> shift;
      int lev = level(node);
      auto bound = bounds(lev);
      nshift[lev] = (nshift[lev] + shift) % (bound.second - bound.first + 1);
      if (type == 2 && shift != 0) {
        long long cshift = 2 * shift;
        int i = 1;
        for (; (1ll << lev + i) < 1.2e18; i++, cshift *= 2) {
          nshift[lev + i] = (nshift[lev + i] + posmod(cshift, 1ll << lev + i)) %
                            (1ll << lev + i);
        }
      }
    } else {
      cout << node << ' ';
      auto bound = bounds(level(node));
      long long nlevels = bound.second - bound.first + 1;
      long long zeroorig = node - nlevels + nshift[level(node)];
      if (zeroorig < 0) zeroorig += 2 * nlevels;
      long long orig = zeroorig % nlevels + bound.first;
      while (orig / 2) {
        orig /= 2;
        auto bound = bounds(level(orig));
        long long nlevels = bound.second - bound.first + 1;
        cout << (orig - nlevels - nshift[level(orig)] + 2 * nlevels) % nlevels +
                    bound.first
             << ' ';
      }
      cout << '\n';
    }
  }
}

```