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("Ofast")
using namespace std;
long long getLevel(long long l, long long r, long long val) {
  if (val >= l && val <= r) return 0;
  return 1ll + getLevel(r + 1ll, 2ll * r + 2ll, val);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  vector<long long> M(62, 0);
  vector<long long> S(62, 1);
  for (long long i = 1; i < 62; ++i) {
    S[i] = 2ll * S[i - 1];
  }
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long t, x, k;
    cin >> t >> x;
    x--;
    long long l = getLevel(0, 0, x);
    if (t == 3) {
      for (long long j = l; j >= 0; j--) {
        cout << x + 1 << " ";
        if (!x) break;
        long long fir = S[j] - 1;
        long long delta = (x - fir + M[j]) % S[j];
        long long pos = fir + delta;
        long long par = (pos - 1ll) / 2ll;
        fir = S[j - 1] - 1ll;
        delta = (par - fir - M[j - 1] + S[j - 1]) % S[j - 1];
        pos = fir + delta;
        x = pos;
      }
      cout << "\n";
      continue;
    }
    cin >> k;
    k += S[61];
    k %= S[l];
    M[l] = (M[l] + k) % S[l];
    if (t == 2) {
      for (long long j = l + 1; j < 62; ++j) {
        k *= 2ll;
        M[j] = (M[j] + k) % S[j];
      }
    }
  }
}