#include <bits/stdc++.h>
using namespace std;
int linie(long long x) {
  for (int i = 60; i >= 0; --i) {
    if (x >= (1LL << i)) {
      return i;
    }
  }
}
void tl(long long &k, int l) {
  k &= (1LL << l) - 1LL;
  if (k < 0) {
    k += 1LL << l;
  }
}
long long op1[70];
long long op2[70];
int main() {
  int q, t, l;
  long long x, k;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    cin >> t;
    if (t == 1) {
      cin >> x >> k;
      l = linie(x);
      tl(k, l);
      op1[l] += k;
      op1[l] &= (1LL << l) - 1LL;
    }
    if (t == 2) {
      cin >> x >> k;
      l = linie(x);
      tl(k, l);
      op2[l] += k;
      op2[l] &= (1LL << l) - 1LL;
    }
    if (t == 3) {
      cin >> x;
      int poz = linie(x);
      x = (x + op1[poz]) % (1LL << poz) + (1LL << poz);
      while (x > 1 && poz > 0) {
        cout << (x - op1[poz]) % (1LL << poz) + (1LL << poz) << " ";
        x = ((x + op2[poz]) % (1LL << poz) + (1LL << poz)) >> 1LL;
        --poz;
      }
      cout << "1\n";
    }
  }
  return 0;
}

### Prompt

Your task is to create 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;
int linie(long long x) {
  for (int i = 60; i >= 0; --i) {
    if (x >= (1LL << i)) {
      return i;
    }
  }
}
void tl(long long &k, int l) {
  k &= (1LL << l) - 1LL;
  if (k < 0) {
    k += 1LL << l;
  }
}
long long op1[70];
long long op2[70];
int main() {
  int q, t, l;
  long long x, k;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    cin >> t;
    if (t == 1) {
      cin >> x >> k;
      l = linie(x);
      tl(k, l);
      op1[l] += k;
      op1[l] &= (1LL << l) - 1LL;
    }
    if (t == 2) {
      cin >> x >> k;
      l = linie(x);
      tl(k, l);
      op2[l] += k;
      op2[l] &= (1LL << l) - 1LL;
    }
    if (t == 3) {
      cin >> x;
      int poz = linie(x);
      x = (x + op1[poz]) % (1LL << poz) + (1LL << poz);
      while (x > 1 && poz > 0) {
        cout << (x - op1[poz]) % (1LL << poz) + (1LL << poz) << " ";
        x = ((x + op2[poz]) % (1LL << poz) + (1LL << poz)) >> 1LL;
        --poz;
      }
      cout << "1\n";
    }
  }
  return 0;
}

```