#include <bits/stdc++.h>
using namespace std;
long long fexp(int b, int x) {
  long long res = 1;
  while (x > 0) {
    if (x % 2 == 1) res = res * b;
    b = b * b;
    x /= 2;
  }
  return res;
}
long long sh[61];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    long long x;
    cin >> x;
    int lv = log2(x);
    if (t == 1) {
      long long k;
      cin >> k;
      sh[lv] += k;
    } else if (t == 2) {
      long long k;
      cin >> k;
      long long ex = 1;
      for (long long i = 0; i < 61 - lv; ++i) {
        sh[lv + i] += ex * k;
        ex *= 2;
      }
    } else {
      long long st = 1LL << lv;
      while (1) {
        cout << x << " ";
        if (x == 1) {
          cout << '\n';
          break;
        }
        long long now = st + ((x + sh[lv]) & (st - 1));
        long long p_orig = now / 2;
        long long p_now = st / 2 + ((p_orig - sh[lv - 1]) & (st / 2 - 1));
        x = p_now;
        lv--;
        st /= 2;
      }
    }
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to 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;
long long fexp(int b, int x) {
  long long res = 1;
  while (x > 0) {
    if (x % 2 == 1) res = res * b;
    b = b * b;
    x /= 2;
  }
  return res;
}
long long sh[61];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    long long x;
    cin >> x;
    int lv = log2(x);
    if (t == 1) {
      long long k;
      cin >> k;
      sh[lv] += k;
    } else if (t == 2) {
      long long k;
      cin >> k;
      long long ex = 1;
      for (long long i = 0; i < 61 - lv; ++i) {
        sh[lv + i] += ex * k;
        ex *= 2;
      }
    } else {
      long long st = 1LL << lv;
      while (1) {
        cout << x << " ";
        if (x == 1) {
          cout << '\n';
          break;
        }
        long long now = st + ((x + sh[lv]) & (st - 1));
        long long p_orig = now / 2;
        long long p_now = st / 2 + ((p_orig - sh[lv - 1]) & (st / 2 - 1));
        x = p_now;
        lv--;
        st /= 2;
      }
    }
  }
  return 0;
}

```