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 N = 1e6 + 7;
const int MX = 1e9 + 7;
const long long int INF = 1e18 + 9LL;
int q;
long long int pot[100];
long long int node[100];
long long int shift[100];
int level(long long int x) {
  int ans = 0;
  while (x > 1) ++ans, x /= 2;
  return ans;
}
void recover(long long int x, int t) {
  if (t == 0) {
    puts("1");
    return;
  }
  long long int y = x - pot[t];
  y += shift[t];
  y %= pot[t];
  y += pot[t];
  printf("%lld ", y);
  x -= pot[t];
  x += node[t];
  x %= pot[t];
  x += pot[t];
  recover(x / 2, t - 1);
}
int main() {
  pot[0] = 1;
  for (int i = 1; i < 61; ++i) pot[i] = pot[i - 1] + pot[i - 1];
  scanf("%d", &q);
  while (q--) {
    int t;
    long long int x, k;
    scanf("%d %lld", &t, &x);
    if (t == 3) {
      if (x == 1) {
        puts("1");
        continue;
      }
      printf("%lld ", x);
      int t = level(x);
      x -= pot[t];
      x += pot[t] - shift[t] + node[t];
      x %= pot[t];
      x += pot[t];
      recover(x / 2, level(x) - 1);
      continue;
    }
    scanf("%lld", &k);
    if (t == 2) {
      int u = level(x);
      node[u] = (((node[u] + k) % pot[u]) + pot[u]) % pot[u];
    } else {
      int u = level(x);
      shift[u] = (((shift[u] - k) % pot[u]) + pot[u]) % pot[u];
    }
  }
  return 0;
}