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;
const int maxn = 65;
long long int a[maxn];
long long int mod[maxn];
int getcen(long long int x) {
  for (int i = 60; i > 0; i--)
    if (x >= mod[i]) return i;
  return 60;
}
int main() {
  int n;
  mod[1] = 1;
  for (int i = 2; i <= 60; i++) {
    mod[i] = 2 * mod[i - 1];
  }
  scanf("%d", &n);
  while (n--) {
    long long int t, x, k;
    scanf("%lld", &t);
    if (t == 1) {
      scanf("%lld%lld", &x, &k);
      int c = getcen(x);
      a[c] = (a[c] + k) % mod[c];
    }
    if (t == 2) {
      scanf("%lld%lld", &x, &k);
      int c = getcen(x);
      k = k % mod[c];
      for (int i = c; i <= 60; i++) {
        a[i] = (a[i] + k) % mod[i];
        k = k * 2;
      }
    }
    if (t == 3) {
      scanf("%lld", &x);
      int c = getcen(x);
      x = x + a[c];
      if (x >= 2 * mod[c]) x -= mod[c];
      if (x < mod[c]) x += mod[c];
      while (c > 0) {
        long long int t = x - a[c];
        if (t < mod[c]) t = t + mod[c];
        if (t >= 2 * mod[c]) t = t - mod[c];
        if (c == 1)
          printf("%lld\n", t);
        else
          printf("%lld ", t);
        x /= 2;
        c--;
      }
    }
  }
  return 0;
}