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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long start[62];
long long num[62];
long long curr[62];
void shift(int cnt, long long k, long long off) {
  while (off--) {
    k = k * 2 % num[cnt];
  }
  k %= num[cnt];
  curr[cnt] -= k;
  curr[cnt] += num[cnt];
  curr[cnt] %= num[cnt];
  curr[cnt] += start[cnt];
}
long long pos(long long x, long long cnt) {
  long long exp = x - start[cnt];
  long long org = x - curr[cnt];
  if (org >= 0)
    return start[cnt] + org;
  else
    return start[cnt] + num[cnt] + org;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  for (int i = 0; i < 62; i++) {
    start[i] = 1LL << i;
    curr[i] = start[i];
  }
  for (int i = 0; i < 61; i++) {
    num[i] = start[i + 1] - start[i];
  }
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      int cnt = 0;
      long long z = x;
      while (z) {
        cnt++;
        z >>= 1;
      }
      cnt--;
      shift(cnt, k, 0);
    } else if (t == 2) {
      long long x, k;
      cin >> x >> k;
      int cnt = 0;
      long long z = x;
      while (z) {
        cnt++;
        z >>= 1;
      }
      cnt--;
      for (int i = cnt; i < 61; i++) {
        shift(i, k, i - cnt);
      }
    } else {
      long long x;
      cin >> x;
      int cnt = 0;
      long long z = x;
      while (z) {
        cnt++;
        z >>= 1;
      }
      cnt--;
      x = pos(x, cnt);
      while (x > 0) {
        int cnt = 0;
        long long z = x;
        while (z) {
          cnt++;
          z >>= 1;
        }
        cnt--;
        z = x - start[cnt] + curr[cnt];
        while (z >= start[cnt] + num[cnt]) {
          z -= num[cnt];
        }
        cout << z << " ";
        x >>= 1;
      }
      cout << '\n';
    }
  }
}