#include <bits/stdc++.h>
using namespace std;
const int fx[] = {+1, -1, +0, +0}, fy[] = {+0, +0, +1, -1};
const long long mod = 1e9 + 7, INF = 1e18, mx = 1e6;
int Case = 1, cas = 0;
long long S[62];
long long Level(long long x) { return 64 - __builtin_clzll(x); }
void update1(long long x, long long k) {
  long long l = Level(x);
  long long M = 1LL << (l - 1);
  S[l] = (S[l] % M + k % M) % M;
}
void update2(long long x, long long k) {
  long long l = Level(x);
  for (int i = l; i <= 61; i++) {
    long long M = 1LL << (i - 1);
    k %= M;
    S[i] = (S[i] + k) % M;
    k *= 2;
  }
}
long long F(long long x, long long s) {
  long long l = Level(x);
  long long M = 1LL << (l - 1);
  long long p = x - M;
  long long pp = p + (S[l] * s);
  pp %= M;
  if (pp < 0) pp += M;
  long long v = M + pp;
  return v;
}
void query(long long x) {
  long long v = F(x, 1);
  while (v) {
    cout << F(v, -1) << " ";
    v /= 2;
  }
  cout << endl;
}
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0);
  ;
  int Q;
  cin >> Q;
  while (Q--) {
    long long t, x, k;
    cin >> t >> x;
    if (t < 3) cin >> k;
    if (t == 1)
      update1(x, k);
    else if (t == 2)
      update2(x, k);
    else
      query(x);
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const int fx[] = {+1, -1, +0, +0}, fy[] = {+0, +0, +1, -1};
const long long mod = 1e9 + 7, INF = 1e18, mx = 1e6;
int Case = 1, cas = 0;
long long S[62];
long long Level(long long x) { return 64 - __builtin_clzll(x); }
void update1(long long x, long long k) {
  long long l = Level(x);
  long long M = 1LL << (l - 1);
  S[l] = (S[l] % M + k % M) % M;
}
void update2(long long x, long long k) {
  long long l = Level(x);
  for (int i = l; i <= 61; i++) {
    long long M = 1LL << (i - 1);
    k %= M;
    S[i] = (S[i] + k) % M;
    k *= 2;
  }
}
long long F(long long x, long long s) {
  long long l = Level(x);
  long long M = 1LL << (l - 1);
  long long p = x - M;
  long long pp = p + (S[l] * s);
  pp %= M;
  if (pp < 0) pp += M;
  long long v = M + pp;
  return v;
}
void query(long long x) {
  long long v = F(x, 1);
  while (v) {
    cout << F(v, -1) << " ";
    v /= 2;
  }
  cout << endl;
}
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0);
  ;
  int Q;
  cin >> Q;
  while (Q--) {
    long long t, x, k;
    cin >> t >> x;
    if (t < 3) cin >> k;
    if (t == 1)
      update1(x, k);
    else if (t == 2)
      update2(x, k);
    else
      query(x);
  }
  return 0;
}

```