#include <bits/stdc++.h>
const int maxn = 61;
long long shift[maxn + 5], num[maxn + 5];
int q;
int get(long long x) {
  int cnt = -1;
  while (x) {
    x = x >> 1;
    cnt++;
  }
  return cnt;
}
void add(long long& a, long long v, long long b) {
  a += v;
  a %= b;
  a = (a + b) % b;
}
int main() {
  for (int i = 0; i < maxn + 1; i++) {
    num[i] = 1LL << i;
    shift[i] = 0;
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int t;
    long long x, k;
    scanf("%d%I64d", &t, &x);
    int h = get(x);
    if (t == 3) {
      x += shift[h];
      if (x > ((1LL << (h + 1)) - 1)) x -= (1LL << h);
      while (x) {
        h = get(x);
        long long k = x;
        k -= shift[h];
        if (k < (1LL << h)) k += (1LL << h);
        printf("%I64d%c", k, x == 1 ? '\n' : ' ');
        x = x / 2;
      }
    } else {
      scanf("%I64d", &k);
      if (t == 1)
        add(shift[h], k, num[h]);
      else {
        for (int j = h; j < 61; j++) {
          add(shift[j], k, num[j]);
          add(k, k, num[j + 1]);
        }
      }
    }
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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>
const int maxn = 61;
long long shift[maxn + 5], num[maxn + 5];
int q;
int get(long long x) {
  int cnt = -1;
  while (x) {
    x = x >> 1;
    cnt++;
  }
  return cnt;
}
void add(long long& a, long long v, long long b) {
  a += v;
  a %= b;
  a = (a + b) % b;
}
int main() {
  for (int i = 0; i < maxn + 1; i++) {
    num[i] = 1LL << i;
    shift[i] = 0;
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int t;
    long long x, k;
    scanf("%d%I64d", &t, &x);
    int h = get(x);
    if (t == 3) {
      x += shift[h];
      if (x > ((1LL << (h + 1)) - 1)) x -= (1LL << h);
      while (x) {
        h = get(x);
        long long k = x;
        k -= shift[h];
        if (k < (1LL << h)) k += (1LL << h);
        printf("%I64d%c", k, x == 1 ? '\n' : ' ');
        x = x / 2;
      }
    } else {
      scanf("%I64d", &k);
      if (t == 1)
        add(shift[h], k, num[h]);
      else {
        for (int j = h; j < 61; j++) {
          add(shift[j], k, num[j]);
          add(k, k, num[j + 1]);
        }
      }
    }
  }
  return 0;
}

```