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 int maxn = 1000;
long long Move1[maxn];
long long Move2[maxn];
inline void op1(int Lay, long long move, long long x, long long times) {
  if (x != 0) Lay = log2(x) + 1;
  long long Base = 1LL << (Lay - 1);
  if (times == -1)
    move %= Base;
  else
    move = ((move % Base) * (times % Base)) % Base;
  Move1[Lay] = ((Move1[Lay] + move) + Base) % Base;
  Move2[Lay] = ((Move2[Lay] - move) + Base) % Base;
}
inline void op2(long long move, long long x) {
  int Lay = log2(x) + 1;
  for (int i = Lay, j = 0; i <= 65; i++, j++) op1(i, move, 0, 1LL << j);
}
inline void op3(long long x) {
  if (x == 1LL) {
    puts("1");
    return;
  }
  printf("%I64d", x);
  int Lay = log2(x) + 1;
  long long Base = 1LL << (Lay - 1);
  long long pos = (((x - Base) + Move1[Lay]) + Base) % Base + Base;
  while (!(pos == 3LL || pos == 2LL)) {
    if (pos & 1)
      pos = (pos - 1) >> 1;
    else
      pos = pos >> 1;
    Lay = log2(pos) + 1;
    Base = 1LL << (Lay - 1);
    printf(" %I64d", ((pos - Base + Move2[Lay]) + Base) % Base + Base);
  }
  puts(" 1");
}
int main(void) {
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int command;
    long long X, K;
    scanf("%d", &command);
    if (command == 1) {
      scanf("%I64d %I64d", &X, &K);
      op1(-1, K, X, -1);
    } else if (command == 2) {
      scanf("%I64d %I64d", &X, &K);
      op2(K, X);
    } else {
      scanf("%I64d", &X);
      op3(X);
    }
  }
  return 0;
}