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;
template <typename T>
inline int read(T& x) {
  int c, sign = 1;
  x = 0;
  while (!(((c = getchar()) >= '0' && c <= '9') || c == '-'))
    if (c == EOF) return c;
  if (c == '-')
    sign = -1;
  else
    x = x * 10 + c - '0';
  while (((c = getchar()) >= '0' && c <= '9')) x = x * 10 + c - '0';
  x *= sign;
  return 1;
}
int main() {
  long long q, t, x, k;
  long long v[60] = {0}, m[64];
  for (int i = 0; i < (64); ++i) m[i] = (1LL << i);
  read(q);
  while (q--) {
    read(t);
    read(x);
    if (t < 3) read(k);
    long long c = 0;
    while (x > m[c + 1] - 1) ++c;
    if (t == 1) {
      v[c] = (v[c] + k) % m[c];
    } else if (t == 2) {
      for (; c < 60; ++c) {
        v[c] = (v[c] + k) % m[c];
        k <<= 1;
      }
    } else {
      long long p = (x - m[c] + v[c] + m[c]) % m[c];
      for (; c > -1; --c) {
        x = m[c] + ((p - v[c] + m[c]) % m[c]);
        cout << x << " ";
        p >>= 1;
      }
      cout << "\n";
    }
  }
  return 0;
}