#include <bits/stdc++.h>
using namespace std;
long long MOD = 1e9 + 7;
const long long maxn = 2e5 + 5;
long long shift[100];
long long numberOfNodes[100];
long long findLevel(long long x) {
  long long ans = 0;
  for (long long i = 62; i >= 0; i--) {
    long long curr = 1ll << i;
    if (curr & x) {
      ans = i + 1;
      break;
    }
  }
  return ans;
}
void f() {
  long long t;
  cin >> t;
  if (t == 1) {
    long long x, k;
    cin >> x >> k;
    long long temp = findLevel(x);
    shift[temp] = (shift[temp] + k) % (numberOfNodes[temp]);
    if (shift[temp] < 0) shift[temp] += numberOfNodes[temp];
  } else if (t == 2) {
    long long x, k;
    cin >> x >> k;
    long long temp = findLevel(x);
    for (long long i = temp; i <= 61; i++) {
      shift[i] = (shift[i] + k) % (numberOfNodes[i]);
      if (shift[i] < 0) shift[i] += numberOfNodes[i];
      k %= numberOfNodes[i];
      if (k < 0) k += numberOfNodes[i];
      k *= 2;
    }
  } else {
    long long x;
    cin >> x;
    long long temp = findLevel(x);
    cout << x << " ";
    ;
    x += shift[temp];
    while (x >= 2 * numberOfNodes[temp]) x -= numberOfNodes[temp];
    x /= 2;
    long long index = x;
    for (long long i = temp - 1; i >= 1; i--) {
      x = index;
      x -= shift[i];
      while (x < numberOfNodes[i]) x += numberOfNodes[i];
      cout << x << " ";
      ;
      index /= 2;
    }
    cout << "" << endl;
    ;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  ;
  for (long long i = 1; i < 62; ++i) {
    numberOfNodes[i] = (1ll << (i - 1));
  }
  long long q;
  cin >> q;
  while (q--) {
    f();
  }
  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;
long long MOD = 1e9 + 7;
const long long maxn = 2e5 + 5;
long long shift[100];
long long numberOfNodes[100];
long long findLevel(long long x) {
  long long ans = 0;
  for (long long i = 62; i >= 0; i--) {
    long long curr = 1ll << i;
    if (curr & x) {
      ans = i + 1;
      break;
    }
  }
  return ans;
}
void f() {
  long long t;
  cin >> t;
  if (t == 1) {
    long long x, k;
    cin >> x >> k;
    long long temp = findLevel(x);
    shift[temp] = (shift[temp] + k) % (numberOfNodes[temp]);
    if (shift[temp] < 0) shift[temp] += numberOfNodes[temp];
  } else if (t == 2) {
    long long x, k;
    cin >> x >> k;
    long long temp = findLevel(x);
    for (long long i = temp; i <= 61; i++) {
      shift[i] = (shift[i] + k) % (numberOfNodes[i]);
      if (shift[i] < 0) shift[i] += numberOfNodes[i];
      k %= numberOfNodes[i];
      if (k < 0) k += numberOfNodes[i];
      k *= 2;
    }
  } else {
    long long x;
    cin >> x;
    long long temp = findLevel(x);
    cout << x << " ";
    ;
    x += shift[temp];
    while (x >= 2 * numberOfNodes[temp]) x -= numberOfNodes[temp];
    x /= 2;
    long long index = x;
    for (long long i = temp - 1; i >= 1; i--) {
      x = index;
      x -= shift[i];
      while (x < numberOfNodes[i]) x += numberOfNodes[i];
      cout << x << " ";
      ;
      index /= 2;
    }
    cout << "" << endl;
    ;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  ;
  for (long long i = 1; i < 62; ++i) {
    numberOfNodes[i] = (1ll << (i - 1));
  }
  long long q;
  cin >> q;
  while (q--) {
    f();
  }
  return 0;
}

```