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 = 1e5 + 5;
long long shift[61];
long long p2[61];
int level(long long X) {
  for (int i = 0; i < 60; ++i) {
    if (X < p2[i + 1] && X >= p2[i]) return i;
  }
  return 60;
}
long long getorig(long long V, long long l) {
  return p2[l] + ((V - p2[l] - shift[l]) % p2[l] + p2[l]) % p2[l];
}
void op1(long long X, long long K) {
  int l = level(X);
  K %= p2[l];
  if (K < 0) K += p2[l];
  shift[l] += K;
  shift[l] %= p2[l];
}
void op2(long long X, long long K) {
  int l = level(X);
  K %= p2[l];
  if (K < 0) K += p2[l];
  for (int i = l; i < 60; ++i) {
    shift[i] += K;
    shift[i] %= p2[i];
    K *= 2;
  }
}
void op3(long long X, int l) {
  long long P = X - p2[l];
  long long V = (p2[l]) + ((P + shift[l]) % p2[l] + p2[l]) % p2[l];
  for (int i = l; i >= 0; --i) {
    long long Z = getorig(V, i);
    printf("%I64d ", Z);
    V /= 2;
  }
  printf("\n");
}
int main() {
  p2[0] = 1;
  for (int i = 1; i < 61; ++i) p2[i] = 2 * p2[i - 1];
  int N;
  scanf("%d \n", &N);
  for (int i = 0; i < N; ++i) {
    int T;
    scanf("%d ", &T);
    if (T != 3) {
      long long K, X;
      scanf("%I64d %I64d \n", &X, &K);
      if (T == 1) op1(X, K);
      if (T == 2) op2(X, K);
    } else {
      long long X;
      scanf("%I64d \n", &X);
      op3(X, level(X));
    }
  }
}