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;
const int N = 1e5 + 5;
int n, q, t;
long long a[70], b[70];
long long X, k;
long long par(long long x) {
  int l = 1 + log2(x);
  x += (a[l] + b[l]) % (1ll << (l - 1));
  if (x >= (1ll << l))
    x -= (1ll << (l - 1));
  else if (x < (1ll << (l - 1)))
    x += (1ll << (l - 1));
  l--;
  x >>= 1;
  x -= a[l];
  if (x >= (1ll << l))
    x -= (1ll << (l - 1));
  else if (x < (1ll << (l - 1)))
    x += (1ll << (l - 1));
  return x;
}
int main() {
  scanf("%d", &q);
  while (q--) {
    scanf("%d%lld", &t, &X);
    if (t == 1) {
      scanf("%lld", &k);
      int l = 1 + log2(X);
      a[l] += k;
      a[l] %= (1ll << (l - 1));
    } else if (t == 2) {
      scanf("%lld", &k);
      int l = 1 + log2(X);
      b[l] += k;
      b[l] %= (1ll << (l - 1));
    } else {
      while (X != 0) {
        printf("%lld ", X);
        if (X == 1) break;
        X = par(X);
      }
      printf("\n");
    }
  }
}