#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void MIN(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline void MAX(T &a, T b) {
  if (a < b) a = b;
}
const int N = 61;
inline long long cur_level(int x) { return 1LL << (x - 1); }
inline long long total_nodes(int x) { return (1LL << x) - 1; }
inline int get_level(long long node) {
  int ret = 1;
  while (total_nodes(ret) < node) ++ret;
  return ret;
}
int main() {
  long long ch[N];
  memset(ch, 0, sizeof(ch));
  int q;
  scanf("%d\n", &q);
  while (q--) {
    int T;
    long long X, K;
    scanf("%d %I64d", &T, &X);
    if (T == 1 or T == 2) {
      scanf("%I64d", &K);
      int L = get_level(X);
      long long SZ = cur_level(L);
      K %= SZ, K += SZ, K %= SZ;
      if (T == 1) {
        ch[L] += K, ch[L] %= SZ;
      } else if (T == 2) {
        for (; L < N; ++L, K <<= 1, SZ <<= 1) ch[L] += K, ch[L] %= SZ;
      }
    } else if (T == 3) {
      int L = get_level(X);
      long long SZ = cur_level(L);
      while (true) {
        printf("%I64d%c", X, " \n"[L == 1]);
        if (L == 1) break;
        X += ch[L], X %= SZ, X += SZ;
        --L, X >>= 1, SZ >>= 1;
        X -= ch[L], X += SZ, X %= SZ, X += SZ;
      }
    }
  }
}

### Prompt

Your challenge is to write a cpp solution to the following 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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void MIN(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline void MAX(T &a, T b) {
  if (a < b) a = b;
}
const int N = 61;
inline long long cur_level(int x) { return 1LL << (x - 1); }
inline long long total_nodes(int x) { return (1LL << x) - 1; }
inline int get_level(long long node) {
  int ret = 1;
  while (total_nodes(ret) < node) ++ret;
  return ret;
}
int main() {
  long long ch[N];
  memset(ch, 0, sizeof(ch));
  int q;
  scanf("%d\n", &q);
  while (q--) {
    int T;
    long long X, K;
    scanf("%d %I64d", &T, &X);
    if (T == 1 or T == 2) {
      scanf("%I64d", &K);
      int L = get_level(X);
      long long SZ = cur_level(L);
      K %= SZ, K += SZ, K %= SZ;
      if (T == 1) {
        ch[L] += K, ch[L] %= SZ;
      } else if (T == 2) {
        for (; L < N; ++L, K <<= 1, SZ <<= 1) ch[L] += K, ch[L] %= SZ;
      }
    } else if (T == 3) {
      int L = get_level(X);
      long long SZ = cur_level(L);
      while (true) {
        printf("%I64d%c", X, " \n"[L == 1]);
        if (L == 1) break;
        X += ch[L], X %= SZ, X += SZ;
        --L, X >>= 1, SZ >>= 1;
        X -= ch[L], X += SZ, X %= SZ, X += SZ;
      }
    }
  }
}

```