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;
long long int Move[65];
long long int Move2[65];
void op1(int f, long long int x, long long int k, long long int times) {
  if (x != 0) f = log2(x) + 1;
  long long int cnt = 1LL << (f - 1);
  k = ((k % cnt) * (times % cnt)) % cnt;
  Move[f] = (Move[f] + k + cnt) % cnt;
  Move2[f] = (Move2[f] - k + cnt) % cnt;
}
void op2(long long int x, long long int k) {
  int f = log2(x) + 1;
  for (int i = 0; f < 65; i++, f++) op1(f, 0, k, 1LL << i);
}
void op3(long long int x) {
  if (x == 1LL) {
    printf("1\n");
    return;
  }
  printf("%lld", x);
  int f = log2(x) + 1;
  long long int cnt = 1LL << (f - 1);
  long long int pos = ((x - cnt) + Move[f] + cnt) % cnt + cnt;
  while (!(pos == 3LL || pos == 2LL)) {
    if (pos & 1)
      pos = (pos - 1) >> 1;
    else
      pos = pos >> 1;
    f = log2(pos) + 1;
    cnt = 1LL << (f - 1);
    printf("% lld", (pos + (cnt - Move[f])) % cnt + cnt);
  }
  printf(" 1\n");
}
int main() {
  memset(Move, 0, sizeof(Move));
  memset(Move2, 0, sizeof(Move2));
  int t;
  scanf("%d", &t);
  while (t--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      long long int x, k;
      scanf("%lld%lld", &x, &k);
      op1(-1, x, k, 1);
    } else if (type == 2) {
      long long int x, k;
      scanf("%lld%lld", &x, &k);
      op2(x, k);
    } else {
      long long int x;
      scanf("%lld", &x);
      op3(x);
    }
  }
  return 0;
}