#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
const int LVL = 61;
int n;
ll add[LVL], add_f[LVL];
bool read() {
  if (!(cin >> n)) {
    return 0;
  }
  return 1;
}
int get_lvl(ll X) {
  int k = 0;
  while ((1ll << k) <= X) {
    k++;
  }
  k--;
  return k;
}
void get(int lvl, ll X) {
  ll len = 1ll << lvl;
  ll val = X - len;
  val = (val + add[lvl]) % len;
  printf("%I64d ", len + val);
  if (lvl == 0) {
    printf("\n");
    return;
  }
  get(lvl - 1, ((X - len + add_f[lvl]) % len + len) / 2);
}
void solve() {
  memset(add, 0, sizeof(add));
  memset(add_f, 0, sizeof(add_f));
  for (int i = 0; i < int(n); ++i) {
    int t;
    ll X, K;
    scanf("%d %I64d", &t, &X);
    int lvl = get_lvl(X);
    ll len = 1ll << lvl;
    if (t == 3) {
      ll tX = X - len;
      tX += len - add[lvl];
      tX %= len;
      tX += len;
      get(lvl, tX);
      continue;
    }
    scanf("%I64d", &K);
    K %= len;
    K += len;
    K %= len;
    if (t == 1) {
      add[lvl] += len - K;
      add[lvl] %= len;
    } else {
      add_f[lvl] += K;
      add_f[lvl] %= len;
    }
  }
}
int main() {
  while (read()) {
    solve();
  }
}

### Prompt

Generate 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;
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
const int LVL = 61;
int n;
ll add[LVL], add_f[LVL];
bool read() {
  if (!(cin >> n)) {
    return 0;
  }
  return 1;
}
int get_lvl(ll X) {
  int k = 0;
  while ((1ll << k) <= X) {
    k++;
  }
  k--;
  return k;
}
void get(int lvl, ll X) {
  ll len = 1ll << lvl;
  ll val = X - len;
  val = (val + add[lvl]) % len;
  printf("%I64d ", len + val);
  if (lvl == 0) {
    printf("\n");
    return;
  }
  get(lvl - 1, ((X - len + add_f[lvl]) % len + len) / 2);
}
void solve() {
  memset(add, 0, sizeof(add));
  memset(add_f, 0, sizeof(add_f));
  for (int i = 0; i < int(n); ++i) {
    int t;
    ll X, K;
    scanf("%d %I64d", &t, &X);
    int lvl = get_lvl(X);
    ll len = 1ll << lvl;
    if (t == 3) {
      ll tX = X - len;
      tX += len - add[lvl];
      tX %= len;
      tX += len;
      get(lvl, tX);
      continue;
    }
    scanf("%I64d", &K);
    K %= len;
    K += len;
    K %= len;
    if (t == 1) {
      add[lvl] += len - K;
      add[lvl] %= len;
    } else {
      add_f[lvl] += K;
      add_f[lvl] %= len;
    }
  }
}
int main() {
  while (read()) {
    solve();
  }
}

```