#include <bits/stdc++.h>
using namespace std;
long long bin[63];
long long tmp[63];
long long power(long long n) {
  if (tmp[n]) return tmp[n];
  long long res = 1;
  for (long long i = 0; i < n; i++) res *= 2;
  tmp[n] = res;
  return res;
}
long long q_mul(long long a, long long b, long long p) {
  long long sum = 0;
  while (b) {
    if (b & 1) {
      (sum += a) %= p;
    }
    (a <<= 1) %= p;
    b >>= 1;
  }
  return sum;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long q;
  cin >> q;
  while (q--) {
    long long t;
    cin >> t;
    if (t < 3) {
      long long x, k, l = -1;
      cin >> x >> k;
      while (x) {
        l++;
        x /= 2;
      }
      bin[l] += k % (power(l));
      bin[l] %= power(l);
      if (bin[l] < 0) bin[l] += power(l);
      if (t == 2) {
        for (long long i = 1; i + l < 63; i++) {
          bin[i + l] += q_mul(k, power(i), power(i + l));
          bin[i + l] %= power(i + l);
          if (bin[i + l] < 0) bin[i + l] += power(i + l);
        }
      }
    } else {
      long long x, y, l = -1;
      cin >> x;
      cout << x << ' ';
      y = x;
      while (y) {
        l++;
        y /= 2;
      }
      y = x - (power(l));
      y += bin[l];
      y %= power(l);
      while (l) {
        y /= 2;
        l--;
        y -= bin[l];
        if (y < 0) y += power(l);
        cout << (power(l)) + y << ' ';
        y += bin[l];
        y %= power(l);
      }
      cout << endl;
    }
  }
  cout << flush;
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 bin[63];
long long tmp[63];
long long power(long long n) {
  if (tmp[n]) return tmp[n];
  long long res = 1;
  for (long long i = 0; i < n; i++) res *= 2;
  tmp[n] = res;
  return res;
}
long long q_mul(long long a, long long b, long long p) {
  long long sum = 0;
  while (b) {
    if (b & 1) {
      (sum += a) %= p;
    }
    (a <<= 1) %= p;
    b >>= 1;
  }
  return sum;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long q;
  cin >> q;
  while (q--) {
    long long t;
    cin >> t;
    if (t < 3) {
      long long x, k, l = -1;
      cin >> x >> k;
      while (x) {
        l++;
        x /= 2;
      }
      bin[l] += k % (power(l));
      bin[l] %= power(l);
      if (bin[l] < 0) bin[l] += power(l);
      if (t == 2) {
        for (long long i = 1; i + l < 63; i++) {
          bin[i + l] += q_mul(k, power(i), power(i + l));
          bin[i + l] %= power(i + l);
          if (bin[i + l] < 0) bin[i + l] += power(i + l);
        }
      }
    } else {
      long long x, y, l = -1;
      cin >> x;
      cout << x << ' ';
      y = x;
      while (y) {
        l++;
        y /= 2;
      }
      y = x - (power(l));
      y += bin[l];
      y %= power(l);
      while (l) {
        y /= 2;
        l--;
        y -= bin[l];
        if (y < 0) y += power(l);
        cout << (power(l)) + y << ' ';
        y += bin[l];
        y %= power(l);
      }
      cout << endl;
    }
  }
  cout << flush;
  return 0;
}

```