#include <bits/stdc++.h>
using namespace std;
template <class T>
bool cmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool cmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const int N = 64;
int Q;
long long S[N];
int lvl(long long x) { return 63 - __builtin_clzll(x); }
long long where(long long x) {
  int i = lvl(x);
  return (x + S[i]) % (1LL << i);
}
long long who(int i, long long p) {
  p = (p - S[i] + (1LL << i)) % (1LL << i);
  return p + (1LL << i);
}
void shift(int i, long long x) {
  S[i] += x;
  S[i] %= (1LL << i);
}
void print() {
  long long x, p;
  scanf("%lld", &x);
  int i = lvl(x);
  p = (1LL << i) + where(x);
  while (i >= 0) {
    printf("%lld ", who(i, p - (1LL << i)));
    p = p / 2, i--;
  }
  puts("");
}
int main() {
  int qt;
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d", &qt);
    if (qt == 3) {
      print();
    } else {
      long long x, k;
      int i;
      scanf("%lld%lld", &x, &k);
      i = lvl(x);
      k %= (1LL << i);
      k += (1LL << i);
      k %= (1LL << i);
      shift(i, k);
      if (qt == 2) {
        for (i = i + 1; i < N; ++i) {
          k = (k * 2) % (1LL << i);
          shift(i, k);
        }
      }
    }
  }
  return 0;
}

### Prompt

Generate 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>
bool cmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool cmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const int N = 64;
int Q;
long long S[N];
int lvl(long long x) { return 63 - __builtin_clzll(x); }
long long where(long long x) {
  int i = lvl(x);
  return (x + S[i]) % (1LL << i);
}
long long who(int i, long long p) {
  p = (p - S[i] + (1LL << i)) % (1LL << i);
  return p + (1LL << i);
}
void shift(int i, long long x) {
  S[i] += x;
  S[i] %= (1LL << i);
}
void print() {
  long long x, p;
  scanf("%lld", &x);
  int i = lvl(x);
  p = (1LL << i) + where(x);
  while (i >= 0) {
    printf("%lld ", who(i, p - (1LL << i)));
    p = p / 2, i--;
  }
  puts("");
}
int main() {
  int qt;
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d", &qt);
    if (qt == 3) {
      print();
    } else {
      long long x, k;
      int i;
      scanf("%lld%lld", &x, &k);
      i = lvl(x);
      k %= (1LL << i);
      k += (1LL << i);
      k %= (1LL << i);
      shift(i, k);
      if (qt == 2) {
        for (i = i + 1; i < N; ++i) {
          k = (k * 2) % (1LL << i);
          shift(i, k);
        }
      }
    }
  }
  return 0;
}

```