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>
template <typename T>
void read(T& re) {
  re = 0;
  int re2 = 1;
  char ch = '!';
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') re2 = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') {
    re = re * 10 + ch - '0';
    ch = getchar();
  }
  re *= re2;
}
template <typename T, typename... Args>
void read(T& re, Args&... args) {
  read(re);
  read(args...);
}
template <typename T>
void write(T A) {
  if (A < 0) {
    A = -A;
    putchar('-');
  }
  if (A / 10) write(A / 10);
  putchar(A % 10 + '0');
}
const int MAXN = 3000 + 7, mod = 998244353;
int q;
long long move[100];
int main() {
  read(q);
  for (long long i = 1, opt, x, k, now, x2; i <= q; i++) {
    read(opt);
    if (opt == 1) {
      read(x, k);
      x2 = x;
      now = -1;
      while (x2) x2 >>= 1, now++;
      k %= (1ll << now);
      move[now] = (move[now] + k + (1ll << now)) % (1ll << now);
    }
    if (opt == 2) {
      read(x, k);
      x2 = x;
      now = -1;
      while (x2) x2 >>= 1, now++;
      k %= (1ll << now);
      for (int i = now; i <= 60; i++) {
        move[i] = (move[i] + k + (1ll << i)) % (1ll << i);
        k *= 2;
      }
    }
    if (opt == 3) {
      read(x);
      x2 = x;
      now = -1;
      while (x2) x2 >>= 1, now++;
      x = x + move[now];
      while (x > (1ll << (now + 1))) x -= 1ll << now;
      while (now >= 0) {
        if (x - move[now] >= (1ll << (now + 1))) {
          write(x - move[now] - (1ll << now));
          putchar(' ');
        } else if (x - move[now] >= (1ll << now)) {
          write(x - move[now]);
          putchar(' ');
        } else {
          write(x - move[now] + (1ll << now));
          putchar(' ');
        }
        x >>= 1;
        now--;
      }
      putchar('\n');
    }
  }
}