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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007, oo = 1000000009;
void Emsawy() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
int dx[] = {1, -1, 0, 0, 1, -1, 1, -1};
int dy[] = {0, 0, 1, -1, 1, -1, -1, 1};
vector<vector<int> > adj;
const long long N = 63;
long long n, m, k, w;
long long lvl[N];
long long LvL(long long x) { return log2(x); }
long long pow2(long long x) {
  if (x < 0) return 0;
  return (1LL << x);
}
void incr(long long x, long long val, bool op) {
  if (op == 1) {
    lvl[LvL(x)] += val;
    return;
  }
  long long inc = 1;
  for (int i = LvL(x); i < N; i++) {
    lvl[i] += (inc)*val;
    long long s = pow2(i), e = pow2(i + 1) - 1;
    long long inter = (e - s + 1);
    lvl[i] %= inter;
    inc *= 2;
  }
}
long long mood(long long x, long long l) {
  long long s = pow2(l), e = pow2(l + 1) - 1;
  long long inter = (e - s + 1);
  x -= s;
  return ((x % inter) + inter) % inter + s;
}
long long shift(long long x, bool op) {
  if (!op) {
    long long l = LvL(x);
    x += (lvl[l]);
    x = mood(x, l);
    return x;
  }
  long long l = LvL(x);
  x += (lvl[l] * -1);
  x = mood(x, l);
  return x;
}
void queryAns(long long x, int idx = 0) {
  long long tmp = x;
  x = shift(x, 1);
  if (idx) cout << x << " ";
  if (x <= 1) return;
  queryAns(tmp / 2, idx + 1);
}
int main() {
  Emsawy();
  while (cin >> n) {
    long long t, k, x;
    while (n--) {
      cin >> t;
      if (t == 3) {
        cin >> x;
        cout << x << " ";
        queryAns(shift(x, 0));
        cout << endl;
        continue;
      }
      cin >> x >> k;
      incr(x, k, t == 1);
    }
  }
  return 0;
}