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>
#include <bits/stdc++.h>
using namespace std;
const double pi = 3.141592653589793;
const int M = 345678;
long long offset[100];
inline int level(long long n) {
  int c = -1;
  while (n) {
    n >>= 1;
    c++;
  }
  return c;
}
inline long long vertex(long long ind, long long l) {
  long long x = ((offset[l] + ind) % (1LL << l) + (1LL << l)) % (1LL << l);
  return x + (1LL << l);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> q;
  while (q--) {
    int t;
    long long x, k;
    cin >> t;
    if (t == 1) {
      cin >> x >> k;
      long long l = level(x);
      offset[l] -= k;
    } else if (t == 2) {
      long long d = 1;
      cin >> x >> k;
      long long l = level(x);
      k %= (1LL << l);
      for (long long i = l; i <= 61; i++) {
        offset[i] -= d * k;
        d <<= 1;
      }
    } else {
      cin >> x;
      long long l = level(x);
      long long ind = x - (1LL << l) - offset[l];
      for (long long i = l; i >= 0; i--) {
        cout << vertex(ind, i) << ' ';
        ind >>= 1;
      }
      cout << "\n";
    }
  }
  return 0;
}
