#include <bits/stdc++.h>
using namespace std;
const long long N = 10000;
const long long M = 60;
vector<long long> v;
long long n, first, t, a[N], second, cn, p;
void val(long long first) {
  long long second, z;
  second = first;
  first = upper_bound((v).begin(), (v).end(), first) - v.begin() - 1;
  ;
  second = (second - v[first] + a[first]) % v[first];
  z = second + v[first];
  while (1) {
    second = z - v[first];
    p = (second - a[first] >= 0) ? (v[first] + (second - a[first]))
                                 : (v[first] * 2 - (a[first] - second));
    cout << p << " ";
    if (p == 1) break;
    first--;
    z >>= 1;
  }
}
int main() {
  cin >> n;
  for (long long i = (long long)0; i <= (long long)M; i++)
    v.push_back((1LL << i));
  for (long long i = (long long)1; i <= (long long)n; i++) {
    cin >> t;
    if (t <= 2) {
      cin >> first >> second;
      first = upper_bound((v).begin(), (v).end(), first) - v.begin() - 1;
      ;
      for (long long j = (long long)first; j <= (long long)M; j++) {
        a[j] += (second % v[j]) * (1LL << (j - first));
        if (a[j] < 0) a[j] %= v[j], a[j] += v[j];
        a[j] %= v[j];
        if (t == 1) break;
      }
    } else {
      cn++;
      cin >> first;
      val(first);
      cout << ""
           << "\n";
    }
  }
}

### Prompt

In cpp, your task is to solve the following 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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long N = 10000;
const long long M = 60;
vector<long long> v;
long long n, first, t, a[N], second, cn, p;
void val(long long first) {
  long long second, z;
  second = first;
  first = upper_bound((v).begin(), (v).end(), first) - v.begin() - 1;
  ;
  second = (second - v[first] + a[first]) % v[first];
  z = second + v[first];
  while (1) {
    second = z - v[first];
    p = (second - a[first] >= 0) ? (v[first] + (second - a[first]))
                                 : (v[first] * 2 - (a[first] - second));
    cout << p << " ";
    if (p == 1) break;
    first--;
    z >>= 1;
  }
}
int main() {
  cin >> n;
  for (long long i = (long long)0; i <= (long long)M; i++)
    v.push_back((1LL << i));
  for (long long i = (long long)1; i <= (long long)n; i++) {
    cin >> t;
    if (t <= 2) {
      cin >> first >> second;
      first = upper_bound((v).begin(), (v).end(), first) - v.begin() - 1;
      ;
      for (long long j = (long long)first; j <= (long long)M; j++) {
        a[j] += (second % v[j]) * (1LL << (j - first));
        if (a[j] < 0) a[j] %= v[j], a[j] += v[j];
        a[j] %= v[j];
        if (t == 1) break;
      }
    } else {
      cn++;
      cin >> first;
      val(first);
      cout << ""
           << "\n";
    }
  }
}

```