#include <bits/stdc++.h>
constexpr int MAX_LVL = 63;
using namespace std;
using ll = long long;
int n;
ll off[MAX_LVL];
int get_lvl(ll x) {
  int ans = 0;
  while (x > 0) {
    ++ans;
    x >>= 1;
  }
  return ans - 1;
}
void shift(int lvl, ll k) {
  ll lvl_size = 1LL << lvl;
  k %= lvl_size;
  k = (k + lvl_size) % lvl_size;
  off[lvl] = (off[lvl] + k) % lvl_size;
}
ll snum(int lvl, ll start, ll amt) {
  ll lvl_size = 1LL << lvl;
  amt = (amt % lvl_size + lvl_size) % lvl_size;
  return lvl_size + (start + amt) % lvl_size;
}
int main() {
  scanf(" %d", &n);
  for (int i = 0; i < n; ++i) {
    int typ;
    scanf(" %d", &typ);
    if (typ == 1) {
      ll x, k;
      scanf(" %lld %lld", &x, &k);
      shift(get_lvl(x), k);
    } else if (typ == 2) {
      ll x, k;
      scanf(" %lld %lld", &x, &k);
      for (int lvl = get_lvl(x); lvl < MAX_LVL; ++lvl) {
        shift(lvl, k);
        k *= 2;
      }
    } else if (typ == 3) {
      ll x;
      scanf(" %lld", &x);
      int lvl = get_lvl(x);
      ll cur_pos = snum(lvl, x, off[lvl]);
      while (cur_pos != 1) {
        printf("%lld ", snum(lvl, cur_pos, -off[lvl]));
        cur_pos /= 2;
        lvl -= 1;
      }
      printf("1\n");
    } else {
      assert(false);
    }
  }
  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>
constexpr int MAX_LVL = 63;
using namespace std;
using ll = long long;
int n;
ll off[MAX_LVL];
int get_lvl(ll x) {
  int ans = 0;
  while (x > 0) {
    ++ans;
    x >>= 1;
  }
  return ans - 1;
}
void shift(int lvl, ll k) {
  ll lvl_size = 1LL << lvl;
  k %= lvl_size;
  k = (k + lvl_size) % lvl_size;
  off[lvl] = (off[lvl] + k) % lvl_size;
}
ll snum(int lvl, ll start, ll amt) {
  ll lvl_size = 1LL << lvl;
  amt = (amt % lvl_size + lvl_size) % lvl_size;
  return lvl_size + (start + amt) % lvl_size;
}
int main() {
  scanf(" %d", &n);
  for (int i = 0; i < n; ++i) {
    int typ;
    scanf(" %d", &typ);
    if (typ == 1) {
      ll x, k;
      scanf(" %lld %lld", &x, &k);
      shift(get_lvl(x), k);
    } else if (typ == 2) {
      ll x, k;
      scanf(" %lld %lld", &x, &k);
      for (int lvl = get_lvl(x); lvl < MAX_LVL; ++lvl) {
        shift(lvl, k);
        k *= 2;
      }
    } else if (typ == 3) {
      ll x;
      scanf(" %lld", &x);
      int lvl = get_lvl(x);
      ll cur_pos = snum(lvl, x, off[lvl]);
      while (cur_pos != 1) {
        printf("%lld ", snum(lvl, cur_pos, -off[lvl]));
        cur_pos /= 2;
        lvl -= 1;
      }
      printf("1\n");
    } else {
      assert(false);
    }
  }
  return 0;
}

```