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: Runtime Error)

//#include <bits/stdc++.h>
//#include <stdio.h>
#include<iostream>
#include<cstdio>
#include<bitset>
#include<algorithm>
#include<vector>
#include<list>
#include<queue>
#include<stack>
#include<string>
#include<string.h>
#include<cmath>
#include<utility>
#include<functional>
#include<map>
#include<set>
#include<cctype>
#include<fstream>


#define FOR(i, a, b) for( int i=(a);i<=(b);i++)
#define RFOR(i, a, b) for( int i=(a);i>=(b);i--)
#define LFOR(i, a, b) for( long long int i=(a);i<=(b);i++)
#define LRFOR(i, a, b) for(long long int i=(a);i>=(b);i--)
#define MOD 1000000007
#define INF 1000000000 //2000000000
#define LLINF 1000000000000000000 //9000000000000000000
#define PI 3.14159265358979
#define MAXI 7500000

using namespace std;
typedef long long int ll;
typedef pair< long long int, long long  int> P;

int dy[5] = { 0,0,1,-1,0 };
int dx[5] = { 1,-1,0,0 ,0 };



int main(void) {
	long long int n, x;
	long long int count = 0;
	string s[51];

	cin >> n >> x;

	s[0] = "o";
	FOR(i, 1, n) {
		s[i] = "x" + s[i - 1] + "o" + s[i - 1] + "x";
	}
	/*FOR(i, 0, n) {
		cout << s[i] << endl;
	}*/
	LFOR(i, 0, x-1) {
		if (s[n][i] == 'o') {
			count++;
		}
	}
	cout << count << endl;

	return 0;
}