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;
long long mx, shift[61];
long long GETLEVEL(long long x) { return (64 - (__builtin_clzll(x))); }
void SHIFT(long long lev, long long k) {
  mx = (1ll << (lev - 1));
  k = k % mx;
  shift[lev] += k;
  if (shift[lev] >= mx) shift[lev] -= mx;
  if (shift[lev] <= -mx) shift[lev] += mx;
}
long long GETNODE(long long lev, long long org) {
  long long x, m, prv, cur;
  x = org;
  m = (1ll << (lev - 1));
  prv = x - m;
  cur = (((prv + (-1ll * shift[lev])) % m) + m) % m;
  x = m + cur;
  return x;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long q;
  cin >> q;
  long long t, x, k, lev, s, m, par, prv, cur, org;
  for (long long i = 0; i < 61; i++) shift[i] = 0;
  for (long long i = 0; i < q; i++) {
    cin >> t;
    if (t == 1) {
      cin >> x >> k;
      lev = GETLEVEL(x);
      SHIFT(lev, k);
    } else if (t == 2) {
      cin >> x >> k;
      lev = GETLEVEL(x);
      m = 0;
      for (long long i = lev; i < 61; i++) {
        s = (1ll << m) * k;
        SHIFT(i, s);
        m++;
      }
    } else {
      cin >> x;
      lev = GETLEVEL(x);
      cout << x << " ";
      m = (1ll << (lev - 1));
      prv = x - m;
      cur = (((prv + shift[lev]) % m) + m) % m;
      x = m + cur;
      x /= 2;
      org = x;
      lev--;
      while (lev > 0) {
        cout << GETNODE(lev, org) << " ";
        org /= 2;
        lev--;
      }
      cout << "\n";
    }
  }
  return 0;
}