#include <bits/stdc++.h>
using namespace std;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
const long long mod = 1e9 + 7;
const int maxn = 1e6 + 5;
long long getd(long long x) {
  long long d = 0;
  long long cur = 2;
  while (cur <= x) {
    cur = cur << 1LL;
    d++;
  }
  return d;
}
long long rot[maxn];
int q;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> q;
  assert(getd(1) == 0);
  assert(getd(5) == 2);
  while (q--) {
    int ty;
    cin >> ty;
    long long x;
    cin >> x;
    long long d = getd(x);
    long long mod = 1LL << d;
    if (ty == 3) {
      long long val = x;
      long long idx = val - mod;
      long long og_idx = (((idx + rot[d]) % mod) + mod) % mod;
      long long og_val = mod + og_idx;
      cout << val << " ";
      while (og_val > 1) {
        og_val = og_val / 2;
        long long d = getd(og_val);
        long long mod = 1LL << d;
        long long idx = og_val - mod;
        long long val = ((idx - rot[d] + mod) % mod) + mod;
        cout << val << " ";
      }
      cout << "\n";
    } else if (ty == 1) {
      long long k;
      cin >> k;
      rot[d] += k;
      rot[d] %= mod;
      rot[d] += mod;
      rot[d] %= mod;
    } else if (ty == 2) {
      long long k;
      cin >> k;
      for (long long j = d; j <= 61; j++) {
        long long mod = 1LL << j;
        long long totr = k;
        for (int w = d; w < j; w++) {
          totr *= 2LL;
          totr %= mod;
        }
        rot[j] += totr;
        rot[j] %= mod;
        rot[j] += mod;
        rot[j] %= mod;
      }
    }
  }
  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 <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
const long long mod = 1e9 + 7;
const int maxn = 1e6 + 5;
long long getd(long long x) {
  long long d = 0;
  long long cur = 2;
  while (cur <= x) {
    cur = cur << 1LL;
    d++;
  }
  return d;
}
long long rot[maxn];
int q;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> q;
  assert(getd(1) == 0);
  assert(getd(5) == 2);
  while (q--) {
    int ty;
    cin >> ty;
    long long x;
    cin >> x;
    long long d = getd(x);
    long long mod = 1LL << d;
    if (ty == 3) {
      long long val = x;
      long long idx = val - mod;
      long long og_idx = (((idx + rot[d]) % mod) + mod) % mod;
      long long og_val = mod + og_idx;
      cout << val << " ";
      while (og_val > 1) {
        og_val = og_val / 2;
        long long d = getd(og_val);
        long long mod = 1LL << d;
        long long idx = og_val - mod;
        long long val = ((idx - rot[d] + mod) % mod) + mod;
        cout << val << " ";
      }
      cout << "\n";
    } else if (ty == 1) {
      long long k;
      cin >> k;
      rot[d] += k;
      rot[d] %= mod;
      rot[d] += mod;
      rot[d] %= mod;
    } else if (ty == 2) {
      long long k;
      cin >> k;
      for (long long j = d; j <= 61; j++) {
        long long mod = 1LL << j;
        long long totr = k;
        for (int w = d; w < j; w++) {
          totr *= 2LL;
          totr %= mod;
        }
        rot[j] += totr;
        rot[j] %= mod;
        rot[j] += mod;
        rot[j] %= mod;
      }
    }
  }
  return 0;
}

```