#include <bits/stdc++.h>
using namespace std;
const int N = 61;
long long sh[N], sz[N], f[N];
int find(long long x) {
  int l = 1, r = N - 1, mid;
  while (l < r) {
    mid = (l + r + 1) / 2;
    if (f[mid] <= x)
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
void go(int lev, long long x) {
  long long g = x - f[lev] + sh[lev] + sz[lev];
  g = (g + sz[lev]) % sz[lev];
  g += f[lev];
  while (lev) {
    x = (g + sz[lev] - sh[lev]) % sz[lev];
    x += f[lev];
    printf("%lld ", x);
    g >>= 1;
    --lev;
  }
  putchar('\n');
}
int main() {
  sz[1] = 1, f[1] = 1;
  for (int i = 2; i < N; ++i) {
    sz[i] = sz[i - 1] * 2;
    f[i] = f[i - 1] * 2;
  }
  int Q;
  scanf("%d", &Q);
  int type, lev;
  long long X, K;
  int i;
  while (Q--) {
    scanf("%d%lld", &type, &X);
    lev = find(X);
    if (type == 1) {
      scanf("%lld", &K);
      K %= sz[lev];
      K = (K + sz[lev]) % sz[lev];
      sh[lev] = (sh[lev] + K) % sz[lev];
    } else if (type == 2) {
      scanf("%lld", &K);
      for (i = lev; i < N; ++i) {
        K %= sz[i];
        K = (K + sz[i]) % sz[i];
        sh[i] = (sh[i] + K) % sz[i];
        K <<= 1;
      }
    } else {
      assert(type == 3);
      assert(lev > 0);
      go(lev, X);
    }
  }
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
const int N = 61;
long long sh[N], sz[N], f[N];
int find(long long x) {
  int l = 1, r = N - 1, mid;
  while (l < r) {
    mid = (l + r + 1) / 2;
    if (f[mid] <= x)
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
void go(int lev, long long x) {
  long long g = x - f[lev] + sh[lev] + sz[lev];
  g = (g + sz[lev]) % sz[lev];
  g += f[lev];
  while (lev) {
    x = (g + sz[lev] - sh[lev]) % sz[lev];
    x += f[lev];
    printf("%lld ", x);
    g >>= 1;
    --lev;
  }
  putchar('\n');
}
int main() {
  sz[1] = 1, f[1] = 1;
  for (int i = 2; i < N; ++i) {
    sz[i] = sz[i - 1] * 2;
    f[i] = f[i - 1] * 2;
  }
  int Q;
  scanf("%d", &Q);
  int type, lev;
  long long X, K;
  int i;
  while (Q--) {
    scanf("%d%lld", &type, &X);
    lev = find(X);
    if (type == 1) {
      scanf("%lld", &K);
      K %= sz[lev];
      K = (K + sz[lev]) % sz[lev];
      sh[lev] = (sh[lev] + K) % sz[lev];
    } else if (type == 2) {
      scanf("%lld", &K);
      for (i = lev; i < N; ++i) {
        K %= sz[i];
        K = (K + sz[i]) % sz[i];
        sh[i] = (sh[i] + K) % sz[i];
        K <<= 1;
      }
    } else {
      assert(type == 3);
      assert(lev > 0);
      go(lev, X);
    }
  }
  return 0;
}

```