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 lvl1(long long x) {
  long long lvl = 0;
  long long p = x;
  while (p != 0) {
    p /= 2;
    lvl++;
  }
  return lvl;
}
long long f(long long x, long long k) {
  long long m = 1;
  long long lvl = lvl1(x);
  lvl--;
  k = (k % (m << lvl) + (m << lvl)) % (m << lvl);
  if (x + k >= (m << (lvl + m))) {
    x = x + k - (m << lvl);
  } else if (x + k < (m << lvl))
    x = x + k + (m << lvl);
  else
    x += k;
  return x;
}
long long f1(long long x, long long k) {
  long long lvl = lvl1(x), m = 1;
  lvl--;
  k = (k % (m << lvl) + (m << lvl)) % (m << lvl);
  if (x - k >= (m << (lvl + m))) {
    x = x - k - (m << lvl);
  } else if (x - k < (m << lvl))
    x = x - k + (m << lvl);
  else
    x -= k;
  return x;
}
unordered_map<long long, long long> s;
int main() {
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      s[lvl1(x)] += k;
    } else if (t == 3) {
      long long x;
      cin >> x;
      cout << x << ' ';
      x = f(x, s[lvl1(x)]);
      x /= 2;
      while (x != 0) {
        cout << f1(x, s[lvl1(x)]) << ' ';
        x /= 2;
      }
      cout << endl;
    } else if (t == 2) {
      long long x, k;
      cin >> x >> k;
      long long lvl = lvl1(x);
      for (int j = lvl; j <= 100; j++) {
        s[j] += k;
        k *= 2;
      }
      int a = 0;
    }
  }
}