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 = 100;
int getLevel(long long x) {
  if (x == 1) return 1;
  return 1 + getLevel(x / 2);
}
int q;
long long cur[maxn];
long long t, x, k;
int main() {
  for (int i = 1; i <= 60; i++) {
    cur[i] = (1ll << (i - 1));
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%lld%lld", &t, &x);
    int l = getLevel(x);
    long long s = (1ll << (l - 1)), L = s, R = s * 2 - 1;
    if (t == 1) {
      scanf("%lld", &k);
      k = (k % s + s) % s;
      cur[l] = (cur[l] - L - k + s) % s + L;
    } else if (t == 2) {
      scanf("%lld", &k);
      k = (k % s + s) % s;
      cur[l] = (cur[l] - L - k + s) % s + L;
      for (int i = l + 1; i <= 60; i++) {
        L *= 2;
        s *= 2;
        k = k * 2 % s;
        cur[i] = (cur[i] - L - k + s) % s + L;
      }
    } else {
      long long now = (x - L - (cur[l] - L) + s) % s + L;
      printf("%lld", x);
      while (now > 1) {
        now /= 2;
        s /= 2;
        L /= 2;
        --l;
        printf(" %lld", (now - L + cur[l] - L) % s + L);
      }
      puts("");
    }
  }
  return 0;
}