#include <bits/stdc++.h>
using namespace std;
using li = long long;
using ld = long double;
void solve(bool);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(20);
  solve(true);
  return 0;
}
void solve(bool __attribute__((unused)) read) {
  li nq;
  cin >> nq;
  vector<li> shift(64);
  auto levelMin = [&](li level) { return (1LL << level); };
  auto levelMax = [&](li level) { return (1LL << (level + 1)) - 1; };
  auto isOnLevel = [&](li x, li level) {
    return x >= levelMin(level) && x <= levelMax(level);
  };
  auto makeMod = [&](li& to, li mod) {
    to %= mod;
    if (to < 0) {
      to += mod;
    }
  };
  auto add = [&](li& to, li x, li mod) {
    to += x;
    makeMod(to, mod);
  };
  for (li i = 0; i < nq; ++i) {
    li t, x, k;
    cin >> t;
    li level = 0;
    auto determine = [&]() {
      while (!isOnLevel(x, level)) ++level;
    };
    if (t == 1) {
      cin >> x >> k;
      determine();
      makeMod(k, 1LL << level);
      add(shift[level], k, 1LL << level);
    } else if (t == 2) {
      cin >> x >> k;
      determine();
      makeMod(k, 1LL << level);
      for (li lev = level; lev < shift.size(); ++lev) {
        add(shift[lev], (k * (1LL << (lev - level))), 1LL << lev);
      }
    } else {
      cin >> x;
      determine();
      x -= levelMin(level);
      x += shift[level];
      makeMod(x, 1LL << level);
      x += levelMin(level);
      while (x) {
        li orig = x;
        x = x - levelMin(level);
        x -= shift[level];
        makeMod(x, 1LL << level);
        x += levelMin(level);
        cout << x << " ";
        x = orig / 2;
        --level;
      }
      cout << "\n";
    }
  }
}

### 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;
using li = long long;
using ld = long double;
void solve(bool);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(20);
  solve(true);
  return 0;
}
void solve(bool __attribute__((unused)) read) {
  li nq;
  cin >> nq;
  vector<li> shift(64);
  auto levelMin = [&](li level) { return (1LL << level); };
  auto levelMax = [&](li level) { return (1LL << (level + 1)) - 1; };
  auto isOnLevel = [&](li x, li level) {
    return x >= levelMin(level) && x <= levelMax(level);
  };
  auto makeMod = [&](li& to, li mod) {
    to %= mod;
    if (to < 0) {
      to += mod;
    }
  };
  auto add = [&](li& to, li x, li mod) {
    to += x;
    makeMod(to, mod);
  };
  for (li i = 0; i < nq; ++i) {
    li t, x, k;
    cin >> t;
    li level = 0;
    auto determine = [&]() {
      while (!isOnLevel(x, level)) ++level;
    };
    if (t == 1) {
      cin >> x >> k;
      determine();
      makeMod(k, 1LL << level);
      add(shift[level], k, 1LL << level);
    } else if (t == 2) {
      cin >> x >> k;
      determine();
      makeMod(k, 1LL << level);
      for (li lev = level; lev < shift.size(); ++lev) {
        add(shift[lev], (k * (1LL << (lev - level))), 1LL << lev);
      }
    } else {
      cin >> x;
      determine();
      x -= levelMin(level);
      x += shift[level];
      makeMod(x, 1LL << level);
      x += levelMin(level);
      while (x) {
        li orig = x;
        x = x - levelMin(level);
        x -= shift[level];
        makeMod(x, 1LL << level);
        x += levelMin(level);
        cout << x << " ";
        x = orig / 2;
        --level;
      }
      cout << "\n";
    }
  }
}

```