#include <bits/stdc++.h>
using namespace std;
void smain();
int main() {
  ios::sync_with_stdio(false);
  smain();
  return 0;
}
const int MAXN = 64 + 3;
long long cnt[MAXN];
long long ori[MAXN];
long long out[MAXN];
void shink(int i) {
  ori[i] = ((ori[i] % cnt[i]) + cnt[i]) % cnt[i];
  out[i] = ((out[i] % cnt[i]) + cnt[i]) % cnt[i];
}
int GetNum(long long g) {
  int ans = 0;
  while (g) ans++, g >>= 1;
  return ans;
}
long long Renew(long long b, long long add) {
  int k = GetNum(b);
  add = (add % (1ll << k - 1) + (1ll << k - 1)) % (1ll << k - 1);
  return (b - (1ll << k - 1) + add) % (1ll << k - 1) + (1ll << k - 1);
}
void smain() {
  int M;
  cin >> M;
  long long d, Len;
  int op, n;
  cnt[0] = 1;
  for (int i = (1); i <= (MAXN - 1); ++i) cnt[i] = 2ll * cnt[i - 1];
  while (M--) {
    cin >> op >> d;
    if (op == 1 || op == 2) cin >> Len;
    n = GetNum(d);
    if (op == 1) {
      out[n] -= Len;
      shink(n);
    }
    if (op == 2) {
      ori[n] += Len;
      shink(n);
    }
    if (op == 3) {
      d = Renew(d, -out[n]);
      while (d) {
        cout << Renew(d, out[n]) << " ";
        d = Renew(d, ori[n]);
        d >>= 1, n--;
      }
      cout << endl;
    }
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
void smain();
int main() {
  ios::sync_with_stdio(false);
  smain();
  return 0;
}
const int MAXN = 64 + 3;
long long cnt[MAXN];
long long ori[MAXN];
long long out[MAXN];
void shink(int i) {
  ori[i] = ((ori[i] % cnt[i]) + cnt[i]) % cnt[i];
  out[i] = ((out[i] % cnt[i]) + cnt[i]) % cnt[i];
}
int GetNum(long long g) {
  int ans = 0;
  while (g) ans++, g >>= 1;
  return ans;
}
long long Renew(long long b, long long add) {
  int k = GetNum(b);
  add = (add % (1ll << k - 1) + (1ll << k - 1)) % (1ll << k - 1);
  return (b - (1ll << k - 1) + add) % (1ll << k - 1) + (1ll << k - 1);
}
void smain() {
  int M;
  cin >> M;
  long long d, Len;
  int op, n;
  cnt[0] = 1;
  for (int i = (1); i <= (MAXN - 1); ++i) cnt[i] = 2ll * cnt[i - 1];
  while (M--) {
    cin >> op >> d;
    if (op == 1 || op == 2) cin >> Len;
    n = GetNum(d);
    if (op == 1) {
      out[n] -= Len;
      shink(n);
    }
    if (op == 2) {
      ori[n] += Len;
      shink(n);
    }
    if (op == 3) {
      d = Renew(d, -out[n]);
      while (d) {
        cout << Renew(d, out[n]) << " ";
        d = Renew(d, ori[n]);
        d >>= 1, n--;
      }
      cout << endl;
    }
  }
}

```