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;
const long double EPS = 1e-9, PI = acos(-1.);
const long long LINF = 0x3f3f3f3f3f3f3f3f, LMOD = (1ll << 60);
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int N = 1e5 + 5;
long long lvl[N];
long long getlvl(long long x) {
  if (x <= 0) return 0;
  long long ans = 1;
  long long k = 1;
  while ((2 * k) <= x) {
    k <<= 1;
    ans++;
  }
  return ans;
}
void add(long long l, long long k) {
  long long v = k;
  for (; l <= 60; l++) {
    lvl[l] += v;
    lvl[l] %= LMOD;
    v *= 2;
  }
}
long long q, t, x, k;
int main() {
  scanf("%lld", &q);
  while (q--) {
    scanf("%lld%lld", &t, &x);
    long long l = getlvl(x);
    if (t == 1) {
      scanf("%lld", &k);
      lvl[l] += k;
      lvl[l] %= LMOD;
    } else if (t == 2) {
      scanf("%lld", &k);
      add(l, k);
    } else if (t == 3) {
      long long re = x + lvl[l];
      re %= (1ll << l);
      while (getlvl(re) > l) re -= (1ll << (l - 1));
      while (getlvl(re) < l) re += (1ll << (l - 1));
      for (int i = l; i >= 1; i--) {
        x = re - lvl[i];
        x %= (1ll << i);
        while (getlvl(x) > i) x -= (1ll << (i - 1));
        while (getlvl(x) < i) x += (1ll << (i - 1));
        printf("%lld ", x);
        re /= 2;
      }
      printf("\n");
    }
  }
  return 0;
}