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;
int q;
long long k1[66], k2[66];
long long LOG[66];
long long get(long long x) {
  int l = 0, r = 60;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (x >= LOG[mid])
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
void go(long long x, int v) {
  printf("%I64d ", x);
  if (!v) return;
  x -= LOG[v];
  x = (x + k1[v] + k2[v]) & (LOG[v] - 1);
  x += LOG[v];
  x = x / 2 - LOG[v - 1];
  x = (x + LOG[v - 1] - k1[v - 1]) & (LOG[v - 1] - 1);
  x += LOG[v - 1];
  go(x, v - 1);
}
void solve() {
  int t;
  scanf("%d", &t);
  if (t < 3) {
    long long x, k;
    scanf("%I64d%I64d", &x, &k);
    long long v = get(x), sum = (1ll << v);
    k %= sum;
    if (t == 1)
      k1[v] = (sum + k1[v] + k) & (sum - 1);
    else
      k2[v] = (sum + k2[v] + k) & (sum - 1);
  } else if (t == 3) {
    long long x;
    scanf("%I64d", &x);
    long long v = get(x);
    go(x, v);
    printf("\n");
  }
}
int main() {
  cin >> q;
  LOG[0] = 1;
  for (int i = 1; i <= 60; ++i) LOG[i] = LOG[i - 1] * 2;
  while (q--) solve();
  return 0;
}