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;
long long q, n, type, first, k, r, poz, a[85], b[85], r1, r2, s[85];
long long getlog(long long first) {
  long long s = 0;
  for (int i = 0; i <= 105; i++) {
    s += (1LL << i);
    if (s >= first) {
      poz = i;
      break;
    }
  }
  return poz + 1;
}
void solve1() {
  scanf("%I64d%I64d", &first, &k);
  poz = getlog(first);
  n = (1LL << poz);
}
long long getpoz(long long n, long long first, long long val) {
  long long f;
  f = n + first + val;
  f = (f & (n - 1));
  if (f == 0) f = n;
  return f;
}
int main() {
  q++;
  q--;
  scanf("%I64d", &q);
  s[1] = 1;
  for (int i = 2; i <= 70; i++) s[i] = s[i - 1] + (1LL << (i - 1));
  while (q--) {
    scanf("%I64d", &type);
    if (type == 1) {
      solve1();
      a[poz] = ((a[poz] + k) & (n - 1));
      continue;
    }
    if (type == 2) {
      solve1();
      b[poz] = ((b[poz] + k) & (n - 1));
      continue;
    }
    scanf("%I64d", &first);
    poz = getlog(first);
    n = (1LL << poz);
    r = getpoz(n, first - s[poz - 1], a[poz]);
    n /= 2;
    do {
      printf("%I64d ", getpoz(n, r, -a[poz]) + s[poz - 1]);
      r = getpoz(n, r, b[poz]);
      r = r + s[poz - 1];
      r /= 2;
      if (r != 1) r -= s[poz - 2];
      n /= 2;
      poz--;
    } while (r != 0);
    printf("\n");
  }
}
