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 = 64;
long long t1[maxn + 5];
long long t2[maxn + 5];
long long base[maxn + 5];
int len(long long n) {
  int ans = 0;
  for (int len = 0; len < 63; len++) {
    if ((1LL << len) & n) {
      ans = len + 1;
    }
  }
  return ans;
}
long long parent(long long x) {
  int idx = len(x);
  long long shift = t1[idx] + t2[idx];
  shift %= base[idx];
  if (shift < 0) shift += base[idx];
  long long new_base = ((x - base[idx] + shift) % base[idx]) + base[idx];
  new_base >>= 1;
  long long shift_p = t1[idx - 1];
  if (shift_p < 0) shift_p += base[idx - 1];
  new_base = new_base - base[idx - 1] - shift_p;
  if (new_base < base[idx - 1]) new_base += base[idx - 1];
  new_base %= base[idx - 1];
  return base[idx - 1] + new_base;
}
int main() {
  int q;
  scanf("%d", &q);
  memset(t1, 0, sizeof t1);
  memset(t2, 0, sizeof t2);
  for (int i = 1; i < maxn; i++) {
    base[i] = 1LL << (i - 1);
  }
  while (q--) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      long long x, k;
      scanf("%I64d %I64d", &x, &k);
      int bits = len(x);
      t1[bits] += k;
      t1[bits] %= base[bits];
    } else if (t == 2) {
      long long x, k;
      scanf("%I64d %I64d", &x, &k);
      int bits = len(x);
      t2[bits] += k;
      t2[bits] %= base[bits];
    } else {
      long long x;
      scanf("%I64d", &x);
      while (x != 1LL) {
        printf("%I64d ", x);
        x = parent(x);
      }
      printf("1\n");
    }
  }
  return 0;
}