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>
#pragma comment(linker, "/STACK:210886422")
using namespace std;
const int MOD = 1000000007;
const int INF = 1000000001;
const long double EPS = 1e-6;
const int HASH_POW = 29;
const long double PI = acos(-1.0);
mt19937_64 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long nextLong(long long l, long long r) {
  uniform_int_distribution<long long> foo(l, r);
  return foo(rnd);
}
double nextDouble(double l, double r) {
  uniform_real_distribution<double> foo(l, r);
  return foo(rnd);
}
double workTime() { return double(clock()) / CLOCKS_PER_SEC; }
void myReturn(int code = 0) { exit(code); }
long long mod[70], mx[70], shift[70];
int findLevel(long long x) {
  int ll = -1, rr = 60;
  while (ll + 1 < rr) {
    int mid = (ll + rr) / 2;
    if (mx[mid] >= x) {
      rr = mid;
    } else {
      ll = mid;
    }
  }
  return rr;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  mod[0] = 1;
  for (int i = 1; i <= 61; ++i) {
    mod[i] = 2 * mod[i - 1];
  }
  mx[0] = 1;
  for (int i = 1; i <= 60; ++i) {
    mx[i] = mx[i - 1] + mod[i];
  }
  int q;
  scanf("%d", &q);
  while (q-- > 0) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      long long x, k;
      scanf("%lld %lld", &x, &k);
      int lev = findLevel(x);
      shift[lev] = (shift[lev] + k) % mod[lev];
      if (shift[lev] < 0) {
        shift[lev] += mod[lev];
      }
    } else if (type == 2) {
      long long x, k;
      scanf("%lld %lld", &x, &k);
      int lev = findLevel(x);
      for (int i = lev; i <= 60; ++i) {
        long long delta = k;
        for (int j = 1; j <= i - lev; ++j) {
          delta = (delta + delta) % mod[i];
        }
        if (delta < 0) {
          delta += mod[i];
        }
        shift[i] = (shift[i] + delta) % mod[i];
      }
    } else {
      long long x;
      scanf("%lld", &x);
      int lev = 0;
      while (mx[lev] < x) {
        ++lev;
      }
      vector<long long> ans;
      ans.push_back(x);
      while (lev > 0) {
        x -= mx[lev - 1] + 1;
        x = (x + shift[lev]) % mod[lev];
        x += mx[lev - 1] + 1;
        x /= 2;
        --lev;
        if (lev > 0) {
          x -= mx[lev - 1] + 1;
          x = (x - shift[lev]) % mod[lev];
          if (x < 0) {
            x += mod[lev];
          }
          x += mx[lev - 1] + 1;
        }
        if (x == 0) {
          x = 1;
        }
        ans.push_back(x);
      }
      for (int i = 0; i < ans.size(); ++i) {
        printf("%lld ", ans[i]);
      }
      printf("\n");
    }
  }
  myReturn();
}