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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
bool MinPlace(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool MaxPlace(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename S, typename T>
ostream &operator<<(ostream &out, const pair<S, T> &p) {
  out << "{" << p.first << ", " << p.second << "}";
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  out << "[";
  for (int i = 0; i < (int)v.size(); i++) {
    out << v[i];
    if (i != (int)v.size() - 1) out << ", ";
  }
  out << "]";
  return out;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int q;
  cin >> q;
  vector<vector<long long int>> s(2, vector<long long int>(62, 0ll));
  while (q--) {
    long long int x;
    int type;
    cin >> type >> x;
    int level = (63 - __builtin_clzll(x));
    if (type == 3) {
      long long int ring = (1ll << level),
                    pos = (x - ring + s[0][level]) % ring;
      while (level >= 0) {
        cout << (((pos + ring - s[0][level]) % ring) + ring) << " ";
        pos = ((pos + s[1][level]) % ring) / 2;
        ring /= 2;
        level--;
      }
      cout << "\n";
    } else {
      long long int k, ring = (1ll << level);
      cin >> k;
      type--;
      k = (k % ring + ring) % ring;
      s[type][level] = (s[type][level] + k) % ring;
    }
  }
};