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>
#include <bits/stdc++.h>
using namespace std;
const int MAXN = -1;
int n, Q;
long long ROT[70], DOS[70];
vector<long long> Ans;
void shiftea(int t, long long S) {
  S = ((S % DOS[t]) + DOS[t]) % DOS[t];
  ROT[t] = (ROT[t] + S) % DOS[t];
}
void solve(long long &X) {
  int t = 0;
  while (DOS[t + 1] <= X) t++;
  Ans.push_back(X);
  long long P;
  X -= DOS[t];
  P = DOS[t] + ((X + ROT[t]) % DOS[t]);
  P /= 2;
  t--;
  P = P - DOS[t];
  P = DOS[t] + ((P - ROT[t] + DOS[t]) % DOS[t]);
  X = P;
}
int main() {
  DOS[0] = 1;
  for (int i = (1); i < (int)(65); i++) DOS[i] = DOS[i - 1] * 2;
  while (scanf("%d", &Q) == 1) {
    for (int i = (0); i < (int)(63); i++) ROT[i] = 0;
    for (int tc = (0); tc < (int)(Q); tc++) {
      int tipo;
      long long X, S;
      scanf("%d %lld", &tipo, &X);
      int t = 0;
      while (DOS[t + 1] <= X) t++;
      switch (tipo) {
        case 1: {
          scanf("%lld", &S);
          shiftea(t, S);
          break;
        }
        case 2: {
          scanf("%lld", &S);
          for (int i = (t); i < (int)(63); i++) {
            shiftea(i, S);
            S = 2 * S % DOS[i + 1];
          }
          break;
        }
        case 3: {
          Ans.clear();
          while (X > 1) solve(X);
          Ans.push_back(1);
          for (int i = (0); i < (int)(Ans.size()); i++) {
            if (i) printf(" ");
            printf("%lld", Ans[i]);
          }
          puts("");
          break;
        }
      }
    }
  }
  return 0;
}
