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>
const long long N = 1000006;
using namespace std;
const long long MOD = 1000000007LL;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T power(T x, T y, long long m = MOD) {
  T ans = 1;
  while (y > 0) {
    if (y & 1LL) ans = (ans * x) % m;
    y >>= 1LL;
    x = (x * x) % m;
  }
  return ans % m;
}
long long shift[66];
inline long long getlev(long long x) {
  long long z = __builtin_clzll(x);
  return 64 - z;
}
inline long long getmod(long long x) { return (1LL << (x - 1)); }
void addshift(long long x, long long k) {
  long long m = getmod(x);
  if (k >= m) k %= m;
  shift[x] += k;
  if (shift[x] >= m) shift[x] -= m;
  if (shift[x] < 0) shift[x] += m;
}
long long getorig(long long V, long long l) {
  long long fval = (1LL << (l - 1));
  long long mod = getmod(l);
  long long P = V - fval;
  long long Z = fval + ((P - shift[l]) % mod + mod) % mod;
  return Z;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> q;
  while (q--) {
    long long t, x, k;
    cin >> t >> x;
    long long l = getlev(x);
    if (t == 1) {
      cin >> k;
      k %= getmod(l);
      addshift(l, k);
    } else if (t == 2) {
      cin >> k;
      k %= getmod(l);
      if (k < 0) k += getmod(l);
      long long curr = 1;
      for (long long i = l; i <= 60; i++) {
        addshift(i, k * curr);
        curr <<= 1;
      }
    } else {
      long long P = x - (1LL << (l - 1));
      long long V = (1LL << (l - 1)) +
                    ((P + shift[l]) % (1LL << (l - 1)) + (1LL << (l - 1))) %
                        (1ll << (l - 1));
      for (long long i = l; i >= 1; i--) {
        long long Z = getorig(V, i);
        cout << Z << " ";
        V >>= 1;
      }
      cout << '\n';
    }
  }
}