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>
#pragma GCC optimize("O3")
using namespace std;
long long n, m, k, p, c;
long long a[((long long)(2e5 + 100))];
set<pair<long long, long long> > v[((long long)(2e5 + 100))];
string s;
long long ch[200];
long long glv(long long x) {
  long long p = 0;
  while (x) {
    p++;
    x /= 2;
  }
  return p - 1;
}
int main() {
  ios::sync_with_stdio(0);
  long long q, qr, x;
  cin >> q;
  while (q--) {
    cin >> qr;
    if (qr == 3) {
      cin >> x;
      long long k = glv(x);
      x &= (1ll << k) - 1;
      x += ch[k];
      x %= (1ll << k);
      for (long long i = k; i >= 0; i--) {
        long long s = (1ll << i);
        cout << (1ll << i) + ((x - ch[i]) % s + s) % s << " ";
        x >>= 1ll;
      }
      cout << endl;
    } else if (qr == 1) {
      cin >> k >> p;
      k = glv(k);
      ch[k] += p;
      ch[k] %= (1ll << k);
      if (ch[k] < 0) ch[k] += (1ll << k);
    } else {
      cin >> k >> p;
      k = glv(k);
      if (p < 0) {
        p %= (1ll << k);
        p += (1ll << k);
      }
      for (long long i = k; i < 62; i++) {
        p &= (1ll << i) - 1;
        ch[i] += p;
        p *= 2;
      }
    }
  }
}