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;
int getd(long long K) {
  int num = 1;
  while (K != 1) {
    num++;
    K = K >> 1;
  }
  return num;
}
long long num[70], A[70], B[70];
void printpath(long long k1, int d) {
  if (d == 0) return;
  long long id = k1 - num[d];
  long long vid = (id - A[d] + num[d]) % num[d] + num[d];
  long long tid = (id + B[d] + num[d]) % num[d] + num[d];
  printf("%lld ", vid);
  printpath(tid / 2, d - 1);
}
int main() {
  num[1] = 1;
  for (int i = 2; i <= 61; i++) num[i] = num[i - 1] * 2;
  int t;
  scanf("%d", &t);
  for (; t; t--) {
    int k1;
    long long k2;
    scanf("%d%lld", &k1, &k2);
    if (k1 == 1) {
      long long k3;
      scanf("%lld", &k3);
      int d = getd(k2);
      A[d] = ((A[d] + k3) % num[d] + num[d]) % num[d];
    } else if (k1 == 2) {
      long long k3;
      scanf("%lld", &k3);
      int d = getd(k2);
      B[d] = ((B[d] + k3) % num[d] + num[d]) % num[d];
    } else {
      int d = getd(k2);
      long long id = k2 - num[d];
      long long tid = (id + A[d] + num[d]) % num[d] + num[d];
      printpath(tid, getd(tid));
      printf("\n");
    }
  }
}
