#include <bits/stdc++.h>
#pragma GCC target("sse4,tune=native")
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;
template <typename T1, typename T2>
inline void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const pair<T, U> &_p) {
  _out << _p.first << ' ' << _p.second;
  return _out;
}
template <typename T, typename U>
inline istream &operator>>(istream &_in, pair<T, U> &_p) {
  _in >> _p.first >> _p.second;
  return _in;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const vector<T> &_v) {
  if (_v.empty()) {
    return _out;
  }
  _out << _v.front();
  for (auto _it = ++_v.begin(); _it != _v.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline istream &operator>>(istream &_in, vector<T> &_v) {
  for (auto &_i : _v) {
    _in >> _i;
  }
  return _in;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const set<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const multiset<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const unordered_set<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const unordered_multiset<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const map<T, U> &_m) {
  if (_m.empty()) {
    return _out;
  }
  _out << '(' << _m.begin()->first << ": " << _m.begin()->second << ')';
  for (auto _it = ++_m.begin(); _it != _m.end(); ++_it) {
    _out << ", (" << _it->first << ": " << _it->second << ')';
  }
  return _out;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const unordered_map<T, U> &_m) {
  if (_m.empty()) {
    return _out;
  }
  _out << '(' << _m.begin()->first << ": " << _m.begin()->second << ')';
  for (auto _it = ++_m.begin(); _it != _m.end(); ++_it) {
    _out << ", (" << _it->first << ": " << _it->second << ')';
  }
  return _out;
}
const string FILENAME = "input";
const int MAXN = 100001;
int q;
long long shift[100];
long long shift1[100];
long long getwho(long long x) {
  int cnt = 0;
  long long x1 = x;
  while (x1 >= 1) {
    cnt++;
    x1 >>= 1;
  }
  long long len = (1LL << (cnt - 1));
  long long pos = x - (1LL << (cnt - 1));
  pos -= shift[cnt];
  pos %= len;
  if (pos < 0) {
    pos += len;
  }
  return pos + (1LL << (cnt - 1));
}
long long getprev(long long x) {
  int cnt = 0;
  long long x1 = x;
  while (x1 >= 1) {
    cnt++;
    x1 >>= 1;
  }
  long long len = (1LL << (cnt - 1));
  long long pos = x - (1LL << (cnt - 1));
  pos += shift1[cnt];
  pos %= len;
  return (pos + (1LL << (cnt - 1))) >> 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> q;
  for (int it = 0; it < q; it++) {
    int t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      int cnt = 0;
      long long x1 = x;
      while (x1 >= 1) {
        cnt++;
        x1 >>= 1;
      }
      shift[cnt] += k;
      shift[cnt] %= (1LL << (cnt - 1));
      if (shift[cnt] < 0) {
        shift[cnt] += (1LL << (cnt - 1));
      }
    } else if (t == 2) {
      long long x, k;
      cin >> x >> k;
      long long x1 = x;
      int cnt = 0;
      while (x1 >= 1) {
        cnt++;
        x1 >>= 1;
      }
      shift1[cnt] += k;
      shift1[cnt] %= (1LL << (cnt - 1));
      if (shift1[cnt] < 0) {
        shift1[cnt] += (1LL << (cnt - 1));
      }
    } else {
      long long x;
      cin >> x;
      long long x1 = x;
      int cnt = 0;
      while (x1 >= 1) {
        cnt++;
        x1 >>= 1;
      }
      long long len = 1LL << (cnt - 1);
      x -= len;
      x = x + shift[cnt];
      if (x >= len) {
        x -= len;
      }
      x += len;
      while (x >= 1) {
        cout << getwho(x) << ' ';
        x = getprev(x);
      }
      cout << '\n';
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#pragma GCC target("sse4,tune=native")
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;
template <typename T1, typename T2>
inline void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const pair<T, U> &_p) {
  _out << _p.first << ' ' << _p.second;
  return _out;
}
template <typename T, typename U>
inline istream &operator>>(istream &_in, pair<T, U> &_p) {
  _in >> _p.first >> _p.second;
  return _in;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const vector<T> &_v) {
  if (_v.empty()) {
    return _out;
  }
  _out << _v.front();
  for (auto _it = ++_v.begin(); _it != _v.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline istream &operator>>(istream &_in, vector<T> &_v) {
  for (auto &_i : _v) {
    _in >> _i;
  }
  return _in;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const set<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const multiset<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const unordered_set<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const unordered_multiset<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const map<T, U> &_m) {
  if (_m.empty()) {
    return _out;
  }
  _out << '(' << _m.begin()->first << ": " << _m.begin()->second << ')';
  for (auto _it = ++_m.begin(); _it != _m.end(); ++_it) {
    _out << ", (" << _it->first << ": " << _it->second << ')';
  }
  return _out;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const unordered_map<T, U> &_m) {
  if (_m.empty()) {
    return _out;
  }
  _out << '(' << _m.begin()->first << ": " << _m.begin()->second << ')';
  for (auto _it = ++_m.begin(); _it != _m.end(); ++_it) {
    _out << ", (" << _it->first << ": " << _it->second << ')';
  }
  return _out;
}
const string FILENAME = "input";
const int MAXN = 100001;
int q;
long long shift[100];
long long shift1[100];
long long getwho(long long x) {
  int cnt = 0;
  long long x1 = x;
  while (x1 >= 1) {
    cnt++;
    x1 >>= 1;
  }
  long long len = (1LL << (cnt - 1));
  long long pos = x - (1LL << (cnt - 1));
  pos -= shift[cnt];
  pos %= len;
  if (pos < 0) {
    pos += len;
  }
  return pos + (1LL << (cnt - 1));
}
long long getprev(long long x) {
  int cnt = 0;
  long long x1 = x;
  while (x1 >= 1) {
    cnt++;
    x1 >>= 1;
  }
  long long len = (1LL << (cnt - 1));
  long long pos = x - (1LL << (cnt - 1));
  pos += shift1[cnt];
  pos %= len;
  return (pos + (1LL << (cnt - 1))) >> 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> q;
  for (int it = 0; it < q; it++) {
    int t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      int cnt = 0;
      long long x1 = x;
      while (x1 >= 1) {
        cnt++;
        x1 >>= 1;
      }
      shift[cnt] += k;
      shift[cnt] %= (1LL << (cnt - 1));
      if (shift[cnt] < 0) {
        shift[cnt] += (1LL << (cnt - 1));
      }
    } else if (t == 2) {
      long long x, k;
      cin >> x >> k;
      long long x1 = x;
      int cnt = 0;
      while (x1 >= 1) {
        cnt++;
        x1 >>= 1;
      }
      shift1[cnt] += k;
      shift1[cnt] %= (1LL << (cnt - 1));
      if (shift1[cnt] < 0) {
        shift1[cnt] += (1LL << (cnt - 1));
      }
    } else {
      long long x;
      cin >> x;
      long long x1 = x;
      int cnt = 0;
      while (x1 >= 1) {
        cnt++;
        x1 >>= 1;
      }
      long long len = 1LL << (cnt - 1);
      x -= len;
      x = x + shift[cnt];
      if (x >= len) {
        x -= len;
      }
      x += len;
      while (x >= 1) {
        cout << getwho(x) << ' ';
        x = getprev(x);
      }
      cout << '\n';
    }
  }
  return 0;
}

```