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>
#pragma warning(disable : 4996)
using namespace std;
const int maxn = 1e6 + 5;
long long level[65];
long long expo[65];
void init() {
  expo[0] = 1;
  for (int i = 1; i < 64; i++) {
    expo[i] = 2 * expo[i - 1];
  }
}
int cal(long long X) {
  int i = 0;
  while (X / 2 != 0) {
    i++;
    X /= 2;
  }
  return i;
}
long long calpos(long long X) {
  int i = cal(X);
  long long a = (X - expo[i]) + level[i];
  long long pos = (a % expo[i] + expo[i]) % expo[i];
  pos += expo[i];
  return pos;
}
long long calvalue(long long pos) {
  int i = cal(pos);
  long long a = (pos - expo[i]) - level[i];
  long long X = (a % expo[i] + expo[i]) % expo[i];
  X += expo[i];
  return X;
}
void query(long long pos) {
  if (pos == 0) return;
  int i = cal(pos);
  long long ans = calvalue(pos);
  cout << ans << ' ';
  query(pos / 2);
}
void Q1(int i, long long K) {
  level[i] += K;
  level[i] = (level[i] % expo[i] + expo[i]) % expo[i];
}
void Q2(int i, long long K) {
  if (i > 63)
    return;
  else {
    K %= expo[i];
    Q1(i, K);
    Q2(i + 1, 2 * K);
  }
}
int main() {
  init();
  int Q;
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    int T;
    cin >> T;
    if (T == 1) {
      long long X;
      long long K;
      cin >> X >> K;
      int p = cal(X);
      Q1(p, K);
    } else if (T == 2) {
      long long X;
      long long K;
      cin >> X >> K;
      int p = cal(X);
      Q2(p, K);
    } else if (T == 3) {
      long long X, pos;
      cin >> X;
      pos = calpos(X);
      query(pos);
      cout << endl;
    }
  }
}