#include <bits/stdc++.h>
using namespace std;
const int64_t logn = 61;
int64_t level(int64_t x) {
  int64_t ans = 0;
  while (x) {
    x /= 2;
    ans++;
  }
  return ans;
}
int64_t sva[logn], sve[logn];
void shift_val(int64_t lvl, int64_t k) {
  sva[lvl - 1] += k;
  sva[lvl - 1] %= 1LL << logn;
}
void shift_ver(int64_t lvl, int64_t k) {
  sve[lvl - 1] += k;
  sve[lvl - 1] %= 1LL << logn;
}
void normalize(int64_t &x, int64_t t) {
  x %= (1LL << t);
  if (x < 0) {
    x += 1LL << t;
  }
  x += 1LL << t;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int64_t q;
  cin >> q;
  while (q--) {
    int64_t t, x;
    cin >> t >> x;
    if (t == 3) {
      vector<int64_t> ans;
      for (int64_t i = logn - 1; i >= 0; i--) {
        if (level(x) > i) {
          ans.push_back(x);
          x += sva[i];
          x += sve[i];
          normalize(x, i);
          x /= 2;
          if (i) {
            x -= sva[i - 1];
            normalize(x, i - 1);
          }
        }
      }
      for (auto it : ans) {
        cout << it << ' ';
      }
      cout << endl;
    } else {
      int64_t k;
      cin >> k;
      if (t == 1) {
        shift_val(level(x), k);
      } else {
        shift_ver(level(x), k);
      }
    }
  }
  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;
const int64_t logn = 61;
int64_t level(int64_t x) {
  int64_t ans = 0;
  while (x) {
    x /= 2;
    ans++;
  }
  return ans;
}
int64_t sva[logn], sve[logn];
void shift_val(int64_t lvl, int64_t k) {
  sva[lvl - 1] += k;
  sva[lvl - 1] %= 1LL << logn;
}
void shift_ver(int64_t lvl, int64_t k) {
  sve[lvl - 1] += k;
  sve[lvl - 1] %= 1LL << logn;
}
void normalize(int64_t &x, int64_t t) {
  x %= (1LL << t);
  if (x < 0) {
    x += 1LL << t;
  }
  x += 1LL << t;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int64_t q;
  cin >> q;
  while (q--) {
    int64_t t, x;
    cin >> t >> x;
    if (t == 3) {
      vector<int64_t> ans;
      for (int64_t i = logn - 1; i >= 0; i--) {
        if (level(x) > i) {
          ans.push_back(x);
          x += sva[i];
          x += sve[i];
          normalize(x, i);
          x /= 2;
          if (i) {
            x -= sva[i - 1];
            normalize(x, i - 1);
          }
        }
      }
      for (auto it : ans) {
        cout << it << ' ';
      }
      cout << endl;
    } else {
      int64_t k;
      cin >> k;
      if (t == 1) {
        shift_val(level(x), k);
      } else {
        shift_ver(level(x), k);
      }
    }
  }
  return 0;
}

```