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>
#include <bits/stdc++.h>
using namespace std;
const pair<int, int> dirs[4] = {make_pair(1, 1), make_pair(1, -1),
                                make_pair(-1, 1), make_pair(-1, -1)},
                     dir[4] = {make_pair(1, 0), make_pair(-1, 0),
                               make_pair(0, 1), make_pair(0, -1)};
const pair<int, int> dirk[8] = {
    make_pair(2, 1), make_pair(2, -1), make_pair(-2, 1), make_pair(-2, -1),
    make_pair(1, 2), make_pair(1, -2), make_pair(-1, 2), make_pair(-1, -2)};
const int inf = 1e9, maxn = 61, mod = 1e9 + 7, N = 3e5 + 20, LOG = 20;
const long double eps = 1e-12, pi = 3.14159265359;
const pair<unsigned long long, unsigned long long> base = make_pair(127, 137);
const long long INF = 1e18;
int q, type;
long long k, x, cnt[maxn], shift[maxn];
void precalc() {
  for (int i = 1; i < maxn; ++i) cnt[i] = (1ll << (i - 1));
}
int lvl(long long x) {
  int lev = 0;
  while (x) {
    x >>= 1ll;
    lev++;
  }
  return lev;
}
void sh(int lev, long long k) {
  shift[lev] = (shift[lev] + k + cnt[lev]) % cnt[lev];
}
int main() {
  precalc();
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &type);
    if (type == 1) {
      scanf("%lld%lld", &x, &k);
      int lev = lvl(x);
      k %= cnt[lev];
      sh(lev, k);
    } else if (type == 2) {
      scanf("%lld%lld", &x, &k);
      int lev = lvl(x);
      for (int i = lev; i < maxn; ++i) {
        k %= cnt[i];
        sh(i, k);
        k <<= 1ll;
      }
    } else {
      scanf("%lld", &x);
      int lev = lvl(x);
      long long curpos = cnt[lev] + (x + shift[lev]) % cnt[lev];
      printf("%lld ", x);
      while (--lev) {
        curpos >>= 1ll;
        x = cnt[lev] + (curpos - shift[lev] + cnt[lev]) % cnt[lev];
        printf("%lld ", x);
      }
      printf("\n");
    }
  }
  return 0;
}
