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 s[60], h, mod;
void solve() {
  long long q;
  cin >> q;
  while (q--) {
    long long t, first, k;
    cin >> t;
    if (t == 1) {
      cin >> first >> k;
      long long r = first;
      h = -1;
      while (r) h++, r /= 2;
      mod = 1LL << h;
      k %= mod;
      if (k < 0) k += mod;
      s[h] += k;
      s[h] %= mod;
    } else if (t == 2) {
      cin >> first >> k;
      long long r = first;
      h = -1;
      while (r) h++, r /= 2;
      mod = 1LL << h;
      k %= mod;
      if (k < 0) k += mod;
      for (long long i = h; i < 60; i++) {
        s[i] += k;
        s[i] %= mod;
        mod = (mod * 2);
        k = (k * 2) % mod;
      }
    } else {
      cin >> first;
      h = -1;
      long long r = first;
      while (r) h++, r /= 2;
      mod = 1LL << h;
      cout << first << " ";
      first = (1LL << h) + (first - (1LL << h) + s[h]) % mod;
      while (first / 2) {
        first /= 2;
        h--;
        mod /= 2;
        long long second = (1LL << h) + (first - (1LL << h) + mod - s[h]) % mod;
        cout << second << " ";
      }
      cout << '\n';
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}