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 a[65];
long long level(long long x) {
  int sum = 0;
  for (; x != 1; x /= 2) {
    sum++;
  }
  return sum;
}
void write(long long a) {
  if (a == 0)
    putchar('0');
  else {
    static char c[20];
    static int c0;
    c0 = 0;
    while (a) c[c0++] = a % 10 + '0', a /= 10;
    while (c0--) putchar(c[c0]);
  }
}
void get_val(long long x, int l) {
  write((x - a[l]) % (1ll << l) + (1ll << l));
  putchar(' ');
  if (x == 1) return;
  get_val(x / 2, l - 1);
}
int main() {
  ios::sync_with_stdio(false);
  int q;
  cin >> q;
  int i;
  for (i = 0; i < q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      x = level(x);
      a[x] += k;
      a[x] %= (1ll << x);
      a[x] += (1ll << x);
      a[x] %= (1ll << x);
    } else if (t == 2) {
      long long x, k;
      cin >> x >> k;
      x = level(x);
      k %= (1ll << x);
      for (; x <= 60; x++) {
        a[x] += k;
        a[x] %= (1ll << x);
        a[x] += (1ll << x);
        a[x] %= (1ll << x);
        k <<= 1;
      }
    } else {
      long long x;
      cin >> x;
      int k = level(x);
      x += a[k];
      x %= (1ll << k);
      x += (1ll << k);
      get_val(x, level(x));
      putchar('\n');
    }
  }
  return 0;
}