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;
vector<long long> v(60), s(60);
int main() {
  for (int i = 0; i < 60; ++i) v[i] = (1ll << i);
  int q;
  scanf("%d", &q);
  while (q--) {
    int t;
    long long x, k = 0;
    scanf("%d%lld", &t, &x);
    if (t <= 2) {
      scanf("%lld\n", &k);
    }
    int h;
    for (h = 0; h < 60 && v[h] <= x; ++h)
      ;
    --h;
    if (t <= 2) {
      for (int i = h, j = t == 1 ? h : 60 - 1; i <= j; ++i) {
        long long mod = v[i], y = (mod - k % mod) % mod;
        s[i] = (s[i] + y) % mod;
        k *= 2;
      }
    } else {
      x = v[h] + (x + v[h] - s[h]) % v[h];
      for (; h > -1; --h) {
        long long mod = v[h];
        printf("%lld ", mod + (x + s[h]) % mod);
        x /= 2;
      }
      printf("\n");
    }
  }
  return 0;
}