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 = int(1e6) + 3;
const int MOD = 1e9 + 7;
long long mul(long long u, long long v, long long mod = MOD) {
  return ((u % mod) * (v % mod)) % mod;
}
long long add(long long u, long long v, long long mod = MOD) {
  return ((u % mod) + (v % mod) + mod) % mod;
}
long long sub(long long u, long long v, long long mod = MOD) {
  return ((u % mod) - (v % mod) + mod) % mod;
}
int Q, T;
long long X, K;
long long shift[70];
int getlog(long long X) {
  int res = -1;
  while (X) {
    res++;
    X /= 2;
  }
  return res;
}
long long getval(long long X, int l, long long sign) {
  long long z = add(shift[l] * sign, X, 1ll << l);
  return z + (1ll << l);
}
int main() {
  cin >> Q;
  while (Q--) {
    cin >> T;
    if (T == 1) {
      cin >> X >> K;
      int l = getlog(X);
      long long val = 1ll << l;
      shift[l] = add(shift[l], K, val);
    } else if (T == 2) {
      cin >> X >> K;
      int l = getlog(X);
      long long val = 1ll << l;
      for (int i = l; i <= 62; i++) {
        shift[i] = add(shift[i], K, val);
        val = val * 2;
        K = K * 2;
      }
    } else {
      cin >> X;
      int l = getlog(X);
      while (X) {
        cout << X << " ";
        long long y = getval(X, l, 1);
        X = y / 2;
        if (X == 0) break;
        X = getval(X, l - 1, -1);
        l--;
      }
      cout << endl;
    }
  }
  return 0;
}