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;
class Solution {
 public:
  long long ONE = 1L;
  long long move[100];
  int lvl(long long x) {
    int lv = 0;
    while (x) {
      lv++;
      x >>= 1;
    }
    return lv - 1;
  }
  void solve(int num, std::istream& in, std::ostream& out) {
    int Q;
    in >> Q;
    for (int i = 0; i < 100; i++) move[i] = 0;
    while (Q--) {
      int kind;
      in >> kind;
      if (kind == 1) {
        long long x, k;
        in >> x >> k;
        if (k == 0) continue;
        int lv = lvl(x);
        long long size = ONE << lv;
        k %= size;
        move[lv] += k;
        move[lv] %= size;
      } else if (kind == 2) {
        long long x, k;
        in >> x >> k;
        if (k == 0) continue;
        int lv = lvl(x);
        long long shift = 1;
        for (int i = lv; i <= 60; i++) {
          long long size = ONE << i;
          k %= size;
          move[i] += (k * shift) % size;
          move[i] %= size;
          shift <<= 1;
        }
      } else if (kind == 3) {
        long long x;
        in >> x;
        int lv = lvl(x);
        long long size = ONE << lv;
        long long poz = (x - size + move[lv]) % size;
        if (poz < 0) poz += size;
        while (lv >= 0) {
          long long shift = (poz - move[lv]) % size;
          if (shift < 0) shift += size;
          long long val = size + shift;
          out << val << " ";
          poz /= 2;
          size >>= 1;
          lv--;
        }
        out << endl;
      }
    }
  }
};
void solve(std::istream& in, std::ostream& out) {
  out << std::setprecision(12);
  Solution solution;
  solution.solve(0, in, out);
}
int main() {
  cin.tie(0);
  istream& in = cin;
  ostream& out = cout;
  solve(in, out);
  return 0;
}