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: Accepted)

#include<stdio.h>

long long int SerchP(int N,long long int X,long long int P[],long long int A[]){
    if (N==0) {
        return 1;
    }
    else if(X==1){
        return 0;
    }
    else if(X<A[N-1]+2){
        return SerchP(N-1,X-1,P,A);
    }
    else if(X==A[N-1]+2){
        return P[N-1]+1;
    }
    else if(X<(2*A[N-1])+3){
        return SerchP(N-1,X-(A[N-1]+2),P,A)+P[N-1]+1;
    }
    else {
        return (2*P[N-1])+1;
    }
}

int main(){
    int N;
    long long int X,A[51],P[51];
    scanf("%d %lld",&N,&X);
    A[0]=1;
    P[0]=1;
    for (int i=1; i<=N; i++) {
        A[i]=(A[i-1]*2)+3;
        P[i]=(P[i-1]*2)+1;
    }
    
    printf("%lld\n",SerchP(N,X,P,A));
    return 0;
}