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 double pi = 2 * acos(0.0);
double inff = 1.0 / 0.0;
long long int x, y, n, i, z, j, k, w, u, q, c, t, m;
pair<long long int, long long int> a[100111];
long long int highest_set(long long int x) {
  x |= (x >> 1);
  x |= (x >> 2);
  x |= (x >> 4);
  x |= (x >> 8);
  x |= (x >> 16);
  x |= (x >> 32);
  return __builtin_ctzll((x + 1) / 2);
}
void print_path(long long int x) {
  bool flag = 0;
  while (1) {
    w = highest_set(x);
    if (!flag) {
      cout << x << " ";
      x = (x + a[w + 1].second) % (1LL << w);
      if (x < 0)
        x += (1LL << (w + 1));
      else
        x += (1LL << (w));
      if (x == 1) break;
      x = x / 2;
      flag = 1;
      continue;
    } else if ((x - a[w + 1].first) % (1LL << (w)) < 0) {
      cout << (x - a[w + 1].first) % (1LL << (w)) + (1LL << (w + 1)) << " ";
    } else
      cout << (x - a[w + 1].first) % (1LL << (w)) + (1LL << w) << " ";
    if (x <= 1) break;
    x = (x - a[w + 1].first + a[w + 1].second) % (1LL << w);
    if (x < 0)
      x += (1LL << (w + 1));
    else
      x += (1LL << (w));
    x = x / 2;
  }
  cout << "\n";
}
void solve() {
  cin >> q;
  if (q == 1 or q == 2) {
    cin >> x >> y;
    w = highest_set(x);
    if (q != 2) a[w + 1].first += y;
    a[w + 1].second += y;
  } else {
    cin >> x;
    print_path(x);
  }
}
int main(void) {
  ios::sync_with_stdio(false), cout.tie(0), cin.tie(0);
  ;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}