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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
#define INF (1<<30)
#define INFLL (1ll<<60)
typedef pair<int, int> P;
typedef pair<int, P> E;
#define MOD (1000000007ll)
#define l_ength size

void mul_mod(ll& a, ll b){
	a *= b;
	a %= MOD;
}

void add_mod(ll& a, ll b){
	b += MOD;
	a += b;
	a %= MOD;
}

ll p[55]={1ll},s[55]={1ll};

int main(void){
	int n,i;
	ll x,ans=0ll,t;
	cin >> n >> x;
	for(i=0; i<n; ++i){
		p[i+1] = 2*p[i]+1;
		s[i+1] = 2*s[i]+3;
	}
	while(n){
		t = (1+s[n])/2;
		if(x==1ll){
			break;
		}else if(x==s[n]){
			ans += p[n];
			break;
		}else if(x<t){
			--x;
		}else{
			ans += p[n-1];
			ans += 1;
			x -= t;
			if(!x){
				break;
			}
		}
		--n;
	}
	cout << ans << endl;
	return 0;
}