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;
const int N = 1010;
long long a[110];
int main() {
  int q;
  cin >> q;
  long long x, k, r, l, mod;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      cin >> x >> k;
      l = log2(x);
      mod = (1ll << l);
      a[l] = (a[l] + k) % mod;
      if (a[l] < 0) a[l] += mod;
    } else if (t == 2) {
      cin >> x >> k;
      l = log2(x);
      while (l < 60) {
        mod = (1ll << l);
        k %= mod;
        a[l] = (a[l] + k) % mod;
        if (a[l] < 0) a[l] += mod;
        l++;
        k *= 2ll;
      }
    } else {
      cin >> x;
      l = log2(x);
      r = x - (1ll << l) + 1;
      mod = (1ll << l);
      r = r + a[l];
      if (r > mod) r = r - mod;
      while (l) {
        mod = (1ll << l);
        k = r - a[l];
        if (k <= 0) k = mod + k;
        cout << (1ll << l) - 1 + k << " ";
        r = (r + 1) / 2;
        l--;
      }
      cout << "1\n";
    }
  }
  return 0;
}