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 RANK = 60;
int Q;
long long T, x, k;
long long det[102], cnt[102];
int get_rank(long long x) {
  int ret = 0;
  while (x) {
    x /= 2;
    ret++;
  }
  return ret;
}
int main() {
  cnt[1] = 1;
  for (int i = 2; i <= RANK + 1; i++) cnt[i] = cnt[i - 1] * 2L;
  scanf("%d", &Q);
  while (Q--) {
    scanf("%lld %lld", &T, &x);
    int rk = get_rank(x);
    if (T == 1) {
      scanf("%lld", &k);
      det[rk] = (det[rk] + k) % cnt[rk];
    }
    if (T == 2) {
      scanf("%lld", &k);
      long long dt = k;
      for (int i = rk; i <= RANK; i++) {
        det[i] = (det[i] + dt) % cnt[i];
        dt = (dt * 2L) % cnt[RANK + 1];
      }
    }
    if (T == 3) {
      long long dt = det[rk];
      dt = (dt % cnt[rk] + cnt[rk]) % cnt[rk];
      long long cur = x + dt;
      if (cur >= cnt[rk + 1]) cur -= cnt[rk];
      printf("%lld ", x);
      cur /= 2;
      while (cur) {
        long long dt = det[--rk];
        dt = (dt % cnt[rk] + cnt[rk]) % cnt[rk];
        long long ans = cur - dt;
        if (ans < cnt[rk]) ans += cnt[rk];
        printf("%lld ", ans);
        cur /= 2;
      }
      printf("\n");
    }
  }
}