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;
template <typename T1, typename T2>
inline void chkmin(T1 &first, T2 second) {
  if (first > second) first = second;
}
template <typename T1, typename T2>
inline void chkmax(T1 &first, T2 second) {
  if (first < second) first = second;
}
const string FILENAME = "input";
const int MAXN = 10000;
int n;
long long w[64];
int get_log(long long X) {
  int ans = 0;
  while (X > 1) {
    X >>= 1;
    ++ans;
  }
  return ans;
}
int main() {
  srand(time(0));
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int t;
    cin >> t;
    if (t == 1) {
      long long X, k;
      cin >> X >> k;
      int gl = get_log(X);
      w[gl] += k;
      w[gl] &= (1LL << gl) - 1;
    } else if (t == 2) {
      long long X, k;
      cin >> X >> k;
      int gl = get_log(X);
      k &= (1LL << gl) - 1;
      for (int i = gl; i < 61; ++i) {
        w[i] += k;
        w[i] &= (1LL << i) - 1;
        k <<= 1;
      }
    } else {
      long long X;
      cin >> X;
      int gl = get_log(X);
      while (X != 1) {
        cout << X << ' ';
        long long p = X - (1LL << gl) + w[gl];
        p &= (1LL << gl) - 1;
        p += (1LL << gl);
        --gl;
        long long Y = (p >> 1);
        Y -= w[gl];
        Y &= (1LL << gl) - 1;
        Y += (1LL << gl);
        X = Y;
      }
      cout << "1\n";
    }
  }
}