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 f[100];
void solve(long long X) {
  cout << X << " ";
  long long k = log2(X);
  if (k == 0) return;
  long long Y = X - (1ll << k);
  Y = (Y + f[k]) % (1ll << k);
  Y += (1ll << k);
  solve(Y / 2);
}
void update(long long k, long long K) {
  f[k] += K;
  f[k] %= (1ll << k);
  if (f[k] < 0) f[k] += (1ll << k);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  memset(f, 0, sizeof f);
  long long Q;
  cin >> Q;
  while (Q--) {
    long long T, X, K;
    cin >> T;
    if (T < 3) {
      cin >> X >> K;
      long long k = 0;
      while (X > 1) X /= 2, k++;
      update(k, K);
      if (T == 1) update(k + 1, -2 * K);
    } else {
      cin >> X;
      solve(X);
      cout << "\n";
    }
  }
}