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 cyc[110];
int main() {
  long long t, i, j, k, l, n, m, q, e = 1;
  string str;
  long long k1, k2;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> m;
  long long x;
  while (m--) {
    cin >> t;
    if (t <= 2) {
      cin >> x >> k;
      long long level = 0, j = 1, op = 0;
      while (level < x) {
        level += j;
        j *= 2;
        op++;
      }
      j /= 2;
      k %= j;
      if (k < 0) k += j;
      if (t == 1)
        cyc[op] += k, cyc[op] %= j;
      else {
        long long jj = k;
        for (i = op; i <= 60; i++) {
          cyc[i] += jj, jj *= 2, cyc[i] %= j, j *= 2;
        }
      }
    } else {
      cin >> x;
      long long level = 0, j = 1, op = 0;
      while (level < x) {
        level += j;
        j *= 2;
        op++;
      }
      j /= 2;
      long long op2 = x - (level - j);
      op2 = op2 + (cyc[op]);
      if (op2 > j) op2 -= j;
      cout << x << ' ';
      op2 += level - j;
      level -= j;
      while (op > 1) {
        op--;
        op2 /= 2;
        j /= 2;
        long long op3 = op2;
        op2 = op3 - (level - j) - cyc[op];
        if (op2 <= 0) op2 += j;
        op2 += level - j;
        cout << op2 << ' ';
        level -= j;
        op2 = op3;
      }
      cout << '\n';
    }
  }
  return 0;
}