In some other world, today is Christmas.

Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:

* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.



For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.

The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?

Constraints

* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.

Input

Input is given from Standard Input in the following format:


N X


Output

Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.

Examples

Input

2 7


Output

4


Input

1 1


Output

0


Input

50 4321098765432109


Output

2160549382716056
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll p[60], a[60];

long long dfs(int level, long long x) {
  if(a[level] <= x) return p[level];
  if(x <= 1) return 0;
  if(a[level-1] + 1 >= x) return dfs(level-1, x - 1);
  return p[level-1] + 1 + dfs(level-1, x - a[level-1] - 2);
}

int main(){
  long long n, x;
  
  cin >> n >> x;
  
  p[0] = 1;
  a[0] = 1;
  for(int i=0;i<59;i++) p[i+1] = p[i] * 2 + 1;
  for(int i=0;i<59;i++) a[i+1] = a[i] * 2 + 3;
  
  cout << dfs(n, x) << endl;
}




