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 <class T>
void chkmax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
void chkmin(T &a, T b) {
  if (a > b) a = b;
}
int q, t;
long long x, k, r[200010];
int f(long long a) {
  long long b = 2;
  int cnt = 0;
  while (b <= a) b <<= 1, cnt++;
  return cnt;
}
int main() {
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &t);
    if (t <= 2) {
      scanf("%I64d%I64d", &x, &k);
      int i = f(x);
      r[i] += k;
      r[i] %= 1ll << i;
      if (t == 1) {
        r[i + 1] -= 2 * k;
        r[i + 1] %= 1ll << (i + 1);
      }
    }
    if (t == 3) {
      scanf("%I64d", &x);
      int i = f(x);
      for (int j = i; j >= 0; j--) {
        printf("%I64d ", x);
        x = (x + r[j]) % (1ll << j) + (1ll << j);
        x >>= 1;
      }
      puts("");
    }
  }
  return 0;
}