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>
#pragma GCC optimize(2)
using namespace std;
const int INF = 1e9 + 7;
const int N = 70;
const int M = 5e6 + 7;
int q;
long long T, X, K;
long long num[N], shift[N];
int Log(long long x) {
  int ans = 0;
  while (x) {
    ans++;
    x >>= 1ll;
  }
  return ans;
}
signed main() {
  ios::sync_with_stdio(false);
  for (int i = 1; i <= 63; i++) num[i] = 1ll << (i - 1);
  cin >> q;
  while (q--) {
    cin >> T;
    if (T == 1) {
      cin >> X >> K;
      shift[Log(X)] += K;
    } else if (T == 2) {
      cin >> X >> K;
      long long now = 1ll;
      for (int dep = Log(X); dep <= 60; dep++) {
        shift[dep] += K * now;
        now <<= 1;
      }
    } else {
      cin >> X;
      for (int dep = Log(X); dep; dep--) {
        cout << X;
        if (dep == 1) {
          cout << "\n";
          break;
        }
        cout << " ";
        long long mov = (shift[dep] % num[dep] + num[dep]) % num[dep];
        long long now = (X - num[dep] + mov) % num[dep] + num[dep];
        X = now / 2;
        mov = ((-shift[dep - 1]) % num[dep - 1] + num[dep - 1]) % num[dep - 1];
        now = (X - num[dep - 1] + mov) % num[dep - 1] + num[dep - 1];
        X = now;
      }
    }
  }
  return 0;
}