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;
template <class s, class d>
ostream &operator<<(ostream &os, pair<s, d> x) {
  return os << "<" << x.f << ", " << x.s << "> ";
}
template <class s>
auto operator<<(ostream &os, s x) ->
    typename enable_if<!is_same<s, string>::value,
                       decltype(x.begin(), os)>::type {
  int o = 0;
  os << "{";
  for (auto c : x) {
    if (o++) os << ", ";
    os << c;
  }
  return os << "} ";
}
void dor() { cerr << endl; }
template <class s, class... t>
void dor(s x, t... y) {
  cerr << x << " ";
  dor(y...);
}
const int M = 61;
long long sh[M], len[M + 1];
int q;
pair<long long, long long> getplace(long long x) {
  long long xx = x;
  long long ind = 0;
  while (xx > 1) {
    ind++;
    xx /= 2;
  }
  long long pos = (x - sh[ind]) % len[ind];
  return make_pair(ind, pos);
}
long long getval(pair<long long, long long> x) {
  return (x.second + sh[x.first]) % len[x.first] + len[x.first];
}
int main() {
  len[0] = 1;
  for (int i = 1; i <= M; i++) {
    len[i] = len[i - 1] * 2;
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    long long a, b, c;
    scanf("%lld%lld", &a, &b);
    pair<long long, long long> x = getplace(b);
    if (a == 3) {
      while (x.first >= 0) {
        printf("%lld ", getval(x));
        x.first--;
        x.second /= 2;
      }
      printf("\n");
      continue;
    }
    scanf("%lld", &c);
    c *= -1;
    c = (c + len[x.first]) % len[x.first];
    if (a == 1) {
      sh[x.first] = (sh[x.first] + len[x.first] + c) % len[x.first];
    } else {
      for (int i = x.first; i < M; i++) {
        sh[i] = (sh[i] + len[i] + c) % len[i];
        c *= 2;
        c %= len[i + 1];
      }
    }
  }
  return 0;
}