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 long long maxn = 1e5 + 123, N = 2e6 + 123;
long long q, a[62], p[62];
long long get(long long x) {
  for (int i = 61; i >= 0; i--)
    if ((1ll << i) & x) return i;
}
void solve(long long x, long long lvl) {
  cout << x << " ";
  if (x == 1) return;
  x = (x + a[lvl]) % p[lvl] + p[lvl];
  x /= 2;
  lvl--;
  x = (x - a[lvl]) % p[lvl] + p[lvl];
  solve(x, lvl);
}
int main() {
  for (int i = 0; i < 62; i++) p[i] = (1ll << i);
  cin >> q;
  while (q--) {
    long long type, x, k;
    cin >> type >> x;
    if (type == 1) {
      cin >> k;
      x = get(x);
      a[x] = (a[x] + k % p[x] + p[x]) % p[x];
    }
    if (type == 2) {
      cin >> k;
      x = get(x);
      for (int i = 0; i + x < 62; i++)
        a[i + x] =
            (a[i + x] + (1ll << i) % p[i + x] * k % p[i + x] + p[i + x]) %
            p[i + x];
    }
    if (type == 3) {
      solve(x, get(x));
      cout << endl;
    }
  }
}
