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;
const int size1 = 4e5;
const long long INF = 1e9;
const long long mod = 1e9 + 7;
const long double eps = 1e-5;
long long get(long long x) {
  long long cnt = 0;
  while (x != 0) {
    cnt++;
    x /= 2;
  }
  return cnt;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> q;
  static long long cycle_shift[100], deg[100];
  deg[0] = 1;
  for (long long i = 1; i <= 61; i++) {
    deg[i] = deg[i - 1] * 2;
  }
  for (int i = 1; i <= q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      long long ur = get(x);
      cycle_shift[ur] += k;
      cycle_shift[ur] =
          ((cycle_shift[ur] % deg[ur - 1]) + deg[ur - 1]) % deg[ur - 1];
    } else {
      if (t == 2) {
        long long x, k;
        cin >> x >> k;
        long long ur = get(x);
        for (long long j = ur; j <= 60; j++) {
          cycle_shift[j] += k;
          cycle_shift[j] =
              ((cycle_shift[j] % deg[j - 1]) + deg[j - 1]) % deg[j - 1];
          k *= 2;
          if (abs(k) > deg[61]) {
            k = (k % deg[61] + deg[61]) % deg[61];
          }
        }
      } else {
        long long x;
        cin >> x;
        long long ur = get(x);
        vector<long long> ans;
        ans.push_back(x);
        long long pos =
            deg[ur - 1] +
            ((x - deg[ur - 1] + cycle_shift[ur]) % deg[ur - 1] + deg[ur - 1]) %
                deg[ur - 1];
        pos /= 2;
        for (long long j = ur - 1; j >= 1; j--) {
          long long cnt = deg[j - 1];
          long long num = pos - cnt;
          long long pol = ((num - cycle_shift[j]) % cnt + cnt) % cnt;
          ans.push_back(cnt + pol);
          pos = ((pol + cycle_shift[j]) % cnt + cnt) % cnt + cnt;
          pos /= 2;
        }
        for (long long el : ans) {
          cout << el << ' ';
        }
        cout << "\n";
      }
    }
  }
  return 0;
}