#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int INF = 0x3f3f3f3f;
const int mxN = 61;
ll level[mxN];
int q;
int getlevel(ll x) {
  int res = 0;
  while ((1LL << res) <= x) {
    res++;
  }
  return res - 1;
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> q;
  ll a, b, c;
  for (int i = 0; i < q; ++i) {
    cin >> a;
    if (a == 1 || a == 2) {
      cin >> b >> c;
      if (a == 1) {
        level[getlevel(b)] += c;
      } else {
        int l = getlevel(b);
        for (int i = l; i < 61; ++i) {
          level[i] = (level[i] + c * (1LL << (i - l))) % (1LL << i);
        }
      }
    } else {
      cin >> b;
      int l = getlevel(b);
      ll out = b - (1LL << l);
      out = (out + level[l]) % (1LL << l);
      if (out < 0) {
        out = out + (1LL << l);
      }
      out = (out + (1LL << l));
      for (int i = l; i >= 0; --i) {
        ll ans = (out - (1LL << i) - level[i]) % (1LL << i);
        if (ans < 0) {
          ans += (1LL << i);
        }
        ans = (ans + (1LL << i));
        cout << ans << " ";
        out /= 2;
      }
      cout << "\n";
    }
  }
  return 0;
}

### Prompt

Generate 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;
using ll = long long;
const int INF = 0x3f3f3f3f;
const int mxN = 61;
ll level[mxN];
int q;
int getlevel(ll x) {
  int res = 0;
  while ((1LL << res) <= x) {
    res++;
  }
  return res - 1;
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> q;
  ll a, b, c;
  for (int i = 0; i < q; ++i) {
    cin >> a;
    if (a == 1 || a == 2) {
      cin >> b >> c;
      if (a == 1) {
        level[getlevel(b)] += c;
      } else {
        int l = getlevel(b);
        for (int i = l; i < 61; ++i) {
          level[i] = (level[i] + c * (1LL << (i - l))) % (1LL << i);
        }
      }
    } else {
      cin >> b;
      int l = getlevel(b);
      ll out = b - (1LL << l);
      out = (out + level[l]) % (1LL << l);
      if (out < 0) {
        out = out + (1LL << l);
      }
      out = (out + (1LL << l));
      for (int i = l; i >= 0; --i) {
        ll ans = (out - (1LL << i) - level[i]) % (1LL << i);
        if (ans < 0) {
          ans += (1LL << i);
        }
        ans = (ans + (1LL << i));
        cout << ans << " ";
        out /= 2;
      }
      cout << "\n";
    }
  }
  return 0;
}

```