Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

2 7

Sample Output 1

4

There are 4 patties in the bottom-most 7 layers of a level-2 burger (BBPPPBPBPPPBB).

Sample Input 2

1 1

Sample Output 2

0

The bottom-most layer of a level-1 burger is a bun.

Sample Input 3

50 4321098765432109

Sample Output 3

2160549382716056

A level-50 burger is rather thick, to the extent that the number of its layers does not fit into a 32-bit integer.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <math.h>

unsigned long long burger(int ,unsigned long long);
unsigned long long lpow(int a, int n);
int main(){
    int n;
    unsigned long long x;
    scanf("%d %llu",&n,&x);
    unsigned long long res;
    res = burger(n,x);
    printf("%llu",res);
    return 0;
    
}
unsigned long long burger(int n,unsigned long long x){
    unsigned long long ret;
    unsigned long long bn;
    bn = lpow(2,n+1) -3;
    if(x == 1){
        ret = 0;
    } else if(x <= bn+1){
        ret = burger(n-1,x-1);
    } else if(x == bn+2){
        ret = (bn+1)/2+1;
    } else if(x <= 2*bn+2){
        ret = (bn+1)/2+1 + burger(n-1,x-(bn+2));
    } else if(x == 2*bn+3){
        unsigned long long an = bn + 2;
        ret = an;
    }
    return ret;
}

unsigned long long lpow(int a, int n){
    int i;
    unsigned long long retn = a;
    for(i=1;i<n;i++){
        retn = retn*a;
    }
    if(n == 0) retn = 1;
    return retn;
}