#include <bits/stdc++.h>
using namespace std;
const long long MAXH = 64 + 3;
long long q, fc[MAXH], sc[MAXH], md[MAXH];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  md[0] = 1;
  for (long long i = 1; i < 64; i++) md[i] = (md[i - 1] << 1);
  cin >> q;
  while (q--) {
    long long t, x, k;
    cin >> t >> x;
    if (t != 3) {
      cin >> k;
      x = (63 - __builtin_clzll(x));
      if (t == 1)
        fc[x] -= k;
      else
        sc[x] -= k;
    } else {
      long long y = (63 - __builtin_clzll(x));
      long long rp = ((((((x - fc[y] - sc[y] - (1ll << y)) % md[y]) + md[y]) %
                        md[y]) +
                       (1ll << y)) >>
                      1),
                f = x;
      vector<long long> v;
      v.push_back(x);
      while (y--) {
        f = ((((rp + fc[y] - (1ll << y)) % md[y]) + md[y]) % md[y]) +
            (1ll << y);
        rp = ((((((f - fc[y] - sc[y] - (1ll << y)) % md[y]) + md[y]) % md[y]) +
               (1ll << y)) >>
              1);
        v.push_back(f);
      }
      for (auto i : v) cout << i << ' ';
      cout << '\n';
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MAXH = 64 + 3;
long long q, fc[MAXH], sc[MAXH], md[MAXH];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  md[0] = 1;
  for (long long i = 1; i < 64; i++) md[i] = (md[i - 1] << 1);
  cin >> q;
  while (q--) {
    long long t, x, k;
    cin >> t >> x;
    if (t != 3) {
      cin >> k;
      x = (63 - __builtin_clzll(x));
      if (t == 1)
        fc[x] -= k;
      else
        sc[x] -= k;
    } else {
      long long y = (63 - __builtin_clzll(x));
      long long rp = ((((((x - fc[y] - sc[y] - (1ll << y)) % md[y]) + md[y]) %
                        md[y]) +
                       (1ll << y)) >>
                      1),
                f = x;
      vector<long long> v;
      v.push_back(x);
      while (y--) {
        f = ((((rp + fc[y] - (1ll << y)) % md[y]) + md[y]) % md[y]) +
            (1ll << y);
        rp = ((((((f - fc[y] - sc[y] - (1ll << y)) % md[y]) + md[y]) % md[y]) +
               (1ll << y)) >>
              1);
        v.push_back(f);
      }
      for (auto i : v) cout << i << ' ';
      cout << '\n';
    }
  }
  return 0;
}

```