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 level(long long x) { return log2(x) + 1; }
long long tr[70];
long long a[70];
int main() {
  long long q, x, k, op;
  for (int i = 1; i <= 64; i++) tr[i] = 1ll << (i - 1);
  scanf("%I64d", &q);
  for (long long i = 1; i <= q; i++) {
    scanf("%I64d", &op);
    if (op == 1) {
      scanf("%I64d%I64d", &x, &k);
      int d = level(x);
      if (k > 0)
        a[d] += k;
      else
        a[d] += tr[d] + k;
      a[d] = (a[d] % tr[d] + tr[d]) % tr[d];
    } else if (op == 2) {
      scanf("%I64d%I64d", &x, &k);
      int d = level(x);
      long long tot = 1;
      for (; d < 64; d++) {
        if (k > 0)
          a[d] += k % tr[d] * tot % tr[d];
        else
          a[d] += (tr[d] + k) % tr[d] * tot % tr[d];
        a[d] = (a[d] % tr[d] + tr[d]) % tr[d];
        tot *= 2;
      }
    } else {
      scanf("%I64d", &x);
      int d = level(x);
      for (int i = d; i >= 1; i--) {
        printf("%I64d ", x);
        x += a[i];
        if (x >= (1ll << i)) x -= (1ll << (i - 1));
        x /= 2;
        x += (tr[i - 1] - a[i - 1]);
        if (x >= (1ll << (i - 1))) x -= (1ll << (i - 2));
      }
      puts("");
    }
  }
  return 0;
}