#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
const long long N = 1e5 + 10;
long long q, cnt[70];
inline long long get(long long x) {
  long long res = 0;
  while (x) {
    x >>= 1LL;
    res++;
  }
  return res;
}
long long qmi(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1LL) res = res * a;
    a = a * a;
    b >>= 1LL;
  }
  return res;
}
signed main() {
  cin >> q;
  while (q--) {
    long long op, x, y, k;
    scanf("%lld %lld", &op, &x);
    y = x;
    x = get(x);
    if (op == 1) {
      scanf("%lld", &k);
      if (k < 0) k += qmi(2, x - 1);
      cnt[x] = (cnt[x] + k) % qmi(2, x - 1);
    } else if (op == 2) {
      scanf("%lld", &k);
      if (k < 0) k += qmi(2, x - 1);
      for (long long i = x; i <= 64; i++, k <<= 1LL)
        cnt[i] = (cnt[i] + k) % qmi(2, i - 1);
    } else {
      while (y) {
        long long h = get(y), num = qmi(2, h - 1);
        printf("%lld ", y);
        y = y + cnt[h];
        y >>= 1LL;
        if (num != 1) y = (y - cnt[h - 1] + num / 2) % (num / 2) + (num / 2);
      }
      puts("");
    }
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
#pragma GCC optimize(2)
using namespace std;
const long long N = 1e5 + 10;
long long q, cnt[70];
inline long long get(long long x) {
  long long res = 0;
  while (x) {
    x >>= 1LL;
    res++;
  }
  return res;
}
long long qmi(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1LL) res = res * a;
    a = a * a;
    b >>= 1LL;
  }
  return res;
}
signed main() {
  cin >> q;
  while (q--) {
    long long op, x, y, k;
    scanf("%lld %lld", &op, &x);
    y = x;
    x = get(x);
    if (op == 1) {
      scanf("%lld", &k);
      if (k < 0) k += qmi(2, x - 1);
      cnt[x] = (cnt[x] + k) % qmi(2, x - 1);
    } else if (op == 2) {
      scanf("%lld", &k);
      if (k < 0) k += qmi(2, x - 1);
      for (long long i = x; i <= 64; i++, k <<= 1LL)
        cnt[i] = (cnt[i] + k) % qmi(2, i - 1);
    } else {
      while (y) {
        long long h = get(y), num = qmi(2, h - 1);
        printf("%lld ", y);
        y = y + cnt[h];
        y >>= 1LL;
        if (num != 1) y = (y - cnt[h - 1] + num / 2) % (num / 2) + (num / 2);
      }
      puts("");
    }
  }
  return 0;
}

```