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>

long long int f(int, long long int);
long long int p[51]={1}, t[51]={1};

int main(void) {
  int i,n,x;
  for (i=1; i<51; i++) {
    p[i] = p[i-1]*2+1;
    t[i] = t[i-1]*2+3;
  }
  scanf("%d%d", &n, &x);
  //  printf(" %d %d\n", n, x);
  //for (i=0; i<51; i++) 
  //  printf(" --[%2d] %lld %lld\n", i, t[i], p[i]);
  //printf("%lld\n",f(n,x));
  return 0;
}

long long int f(int n, long long int x) {
  printf("  %d %lld\n", n, x);
  if (n==0) return 1;
  if (x==1) return 0;
  //long long int m = t[n]/2;
  if (x==t[n])
    return p[n];
  if (x==p[n])
    return p[n-1]+1;
  if (x<p[n])
    return f(n-1, x-1);
  return p[n-1]+f(n-1,x-p[n]-1);
}