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>
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long lg = 62;
long long a[lg];
long long n;
long long get_level(long long x) {
  for (long long i = 0; i < lg; ++i) {
    if ((1LL << (i + 1)) - 1 >= x) return i;
  }
  assert(false);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    long long t, x, k;
    cin >> t;
    if (t == 1) {
      cin >> x >> k;
      x = get_level(x);
      long long kek = (1LL << x);
      k = (k % kek + kek) % kek;
      a[x] += k;
      a[x] = (a[x] + (1LL << x)) % (1LL << x);
    }
    if (t == 2) {
      cin >> x >> k;
      x = get_level(x);
      long long kek = (1LL << x);
      k = (k % kek + kek) % kek;
      for (long long j = x; j < lg; ++j) {
        a[j] += k;
        a[j] = (a[j] + (1LL << j)) % (1LL << j);
        k *= 2;
      }
    }
    if (t == 3) {
      vector<long long> v;
      cin >> x;
      v.push_back(x);
      long long level = get_level(x);
      long long p2 = (1LL << level);
      x = (x + a[level]) % p2 + p2;
      x /= 2;
      while (x > 0) {
        long long level = get_level(x);
        long long p2 = (1LL << level);
        x -= p2;
        long long y = (x - a[level] + p2) % p2;
        v.push_back(y + p2);
        x += p2;
        x /= 2;
      }
      for (auto p : v) {
        cout << p << " ";
      }
      cout << endl;
    }
  }
}
