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;
using ll = long long;
unordered_map<int, ll> shift_nums, shift_trees;
int get_lvl(ll a) {
  int lvl = 0;
  while (a > 0) {
    a >>= 1;
    ++lvl;
  }
  return lvl - 1;
}
ll original_pos(ll x, int lvl) { return x - (1ll << lvl); }
ll mod(ll x, int lvl) {
  ll cnt = 1ll << lvl;
  x %= cnt;
  if (x < 0) x += cnt;
  return x;
}
ll up(ll x, int lvl) {
  if (lvl == 0) return 0;
  ll pos = mod(original_pos(x, lvl) + shift_nums[lvl] + shift_trees[lvl], lvl);
  ll up_pos =
      mod(pos / 2 - shift_nums[lvl - 1] - shift_trees[lvl - 1], lvl - 1) +
      (1ll << (lvl - 1));
  return up_pos;
}
void print(ll x, int lvl) {
  while (lvl != -1) {
    printf("%lld ", x);
    x = up(x, lvl);
    --lvl;
  }
  putchar('\n');
}
int main() {
  int q;
  scanf("%d", &q);
  for (int j = 0; j < q; ++j) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      ll x, k;
      scanf("%lld%lld", &x, &k);
      shift_nums[get_lvl(x)] += k;
    } else if (t == 2) {
      ll x, k;
      scanf("%lld%lld", &x, &k);
      k %= 1ll << get_lvl(x);
      for (int i = get_lvl(x); i <= 61; ++i, k *= 2) {
        shift_trees[i] += k;
      }
    } else {
      ll x;
      scanf("%lld", &x);
      print(x, get_lvl(x));
    }
  }
}