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 INF = 0x3f3f3f3f;
const long long int INFL = 0x3f3f3f3f3f3f3f3f;
const double pi = atan(1.0) * 4.0;
const int N = 1e5 + 500;
const int LOG = 65;
const long long int MAXP = 1LL << 60LL;
long long int rot[LOG];
inline long long int depth(long long int a) {
  long long int ans = 0;
  while (a > 1) {
    ans++;
    a /= 2;
  }
  return ans;
}
inline long long int mod(long long int a, long long int b) {
  a %= b;
  return (a + b) % b;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  long long int q, t, x, k;
  cin >> q;
  for (int i = 0; i < LOG; i++) rot[i] = 0;
  while (q--) {
    cin >> t;
    if (t == 3) {
      cin >> x;
      cout << x;
      long long int d = depth(x);
      long long int fst = 1LL << d;
      long long int rp = fst + mod(x - fst + rot[d], fst);
      while (d > 0) {
        d--;
        fst /= 2;
        rp /= 2;
        cout << " " << fst + mod(rp - fst - rot[d], fst);
      }
      cout << endl;
    } else {
      cin >> x >> k;
      long long int d = depth(x);
      if (t == 1)
        (rot[d] += k) %= MAXP;
      else {
        long long int r = k;
        for (int i = d; i < 65; i++) {
          (rot[i] += r) %= MAXP;
          (r *= 2) %= MAXP;
        }
      }
    }
  }
  return 0;
}