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;
typedef pair<long long int, long long int> P;
const long long int OO = 1e18;
void fast() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
long long int q, x, t, k, sh[int(62)], g = 1, res;
P p[int(62)];
long long int lg(long long int y) {
  for (long long int i = 0; i <= 60; i++)
    if ((1LL << i) > y) return i - 1;
}
void shft(long long int y, long long int w) {
  sh[y] += w;
  bool ch = (sh[y] < 0);
  if (ch) sh[y] = -sh[y];
  sh[y] %= (1LL << y);
  if (ch) sh[y] = -sh[y];
}
int main() {
  for (long long int i = 0; i <= 60; i++) {
    p[i].first = (1LL << i), p[i].second = p[i].first * (2LL) - 1LL;
  }
  scanf("%lld", &q);
  while (q--) {
    scanf("%lld%lld", &t, &x);
    if (t == 1) {
      scanf("%lld", &k);
      shft(lg(x), k);
    } else if (t == 2) {
      scanf("%lld", &k);
      g = k;
      for (int i = lg(x); i <= 60; i++) shft(i, g), g *= 2LL;
    } else {
      g = lg(x);
      x += sh[g];
      if (x < p[g].first) {
        x -= p[g].first;
        x += p[g].second + 1;
      } else if (x > p[g].second) {
        x -= (p[g].second + 1);
        x += p[g].first;
      }
      while (g >= 0) {
        res = x;
        res -= sh[g];
        swap(x, res);
        if (x < p[g].first) {
          x -= p[g].first;
          x += p[g].second + 1;
        } else if (x > p[g].second) {
          x -= (p[g].second + 1);
          x += p[g].first;
        }
        swap(x, res);
        printf("%lld", res);
        if (g != 0)
          printf(" ");
        else
          printf("\n");
        g--;
        x /= 2;
      }
    }
  }
  return 0;
}