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 double eps = 1.0e-10;
const double EPS = 1.0e-4;
const int turn[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
int Q;
int T;
long long level;
long long X, K;
long long lenth, position;
long long shift[70];
void update(long long x, long long K) {
  shift[x] += K;
  shift[x] %= (1LL << (x - 1));
}
void pushdown(long long x, long long K) {
  shift[x] += K;
  shift[x] %= (1LL << (x - 1));
  if (x + 1 < 61) {
    K = (K << 1);
    pushdown(x + 1, K);
  }
}
long long getpos(long long level, long long x) {
  long long l = 1LL << (level - 1);
  long long r = (1LL << level) - 1;
  long long now = x + shift[level];
  if (l <= now && now <= r) {
    return now;
  }
  if (now < l) {
    long long cha = l - now - 1;
    return r - cha;
  } else {
    long long cha = now - r - 1;
    return l + cha;
  }
}
void print(long long level, long long pos) {
  if (level == 1) {
    cout << 1;
    return;
  }
  long long now = pos - shift[level];
  long long l = 1LL << (level - 1);
  long long r = (1LL << level) - 1;
  if (l <= now && now <= r) {
    cout << now << " ";
    print(level - 1, pos / 2);
    return;
  }
  if (now < l) {
    long long cha = l - now - 1;
    cout << r - cha << " ";
  } else {
    long long cha = now - r - 1;
    cout << l + cha << " ";
  }
  print(level - 1, pos / 2);
  return;
}
int main() {
  cin >> Q;
  while (Q--) {
    int T;
    scanf("%d %lld", &T, &X);
    for (long long i = 61; i >= 1; i--) {
      if ((1LL << (i - 1)) & X) {
        if (((1LL << i) & X) == 0) {
          level = i;
          break;
        }
      }
    }
    if (T == 1) {
      scanf("%lld", &K);
      if (X == 1) {
        continue;
      }
      K %= 1LL << (level - 1);
      update(level, K);
    } else if (T == 2) {
      scanf("%lld", &K);
      if (X == 1) {
        continue;
      }
      K %= 1LL << (level - 1);
      pushdown(level, K);
    } else {
      if (X == 1) {
        cout << 1 << endl;
        continue;
      }
      cout << X << " ";
      position = getpos(level, X);
      print(level - 1, position / 2);
      cout << endl;
    }
  }
}