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 long long MAX = 1e5;
long long N, Q, ch, X, Y, level, pos, power[MAX], shift[MAX];
long long LogBase2(long long X) {
  long long res = 0;
  while (X) {
    res++;
    X /= 2;
  }
  return res;
}
long long original(long long X) {
  level = LogBase2(X);
  return X - power[level - 1];
}
long long position(long long X) {
  long long res;
  level = LogBase2(X);
  res = original(X) + shift[level];
  while (res < 0) res += power[level - 1];
  while (res >= power[level - 1]) res -= power[level - 1];
  return power[level - 1] + res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  power[0] = 1;
  for (long long A = 1; A <= 61; A++) power[A] = power[A - 1] + power[A - 1];
  cin >> Q;
  while (Q--) {
    cin >> ch;
    if (ch == 1) {
      cin >> X >> Y;
      X = LogBase2(X);
      if (X > 61) continue;
      shift[X] += Y;
      shift[X] %= power[X - 1];
    } else if (ch == 2) {
      cin >> X >> Y;
      X = LogBase2(X);
      for (long long A = X; A <= 61; A++) {
        shift[A] += (power[A - X] % power[A - 1]) * Y;
        shift[A] %= power[A - 1];
      }
    } else {
      cin >> X;
      level = LogBase2(X);
      Y = position(X);
      while (level) {
        X = (original(Y) - shift[level]);
        while (X < 0) X += power[level - 1];
        while (X >= power[level - 1]) X -= power[level - 1];
        cout << power[level - 1] + X << " ";
        Y /= 2;
        level--;
      }
      cout << "\n";
    }
  }
  return 0;
}