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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
using P = pair<ll,ll>;
const ll mod = 1e9+7;
#define ALL(x) (x).begin(),(x).end()
#define pow(x,y) modpow(x,y)
#define REP(i,n) for(ll (i)=0;(i)<(n);(i)++)

template <class T = int>T in(){
    T x;
    cin >> x;
    return (x);
}

signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout<<fixed<<setprecision(10);
    ll N = in<ll>(),X = in<ll>();
    vector<ll> tpow(N+3);
    tpow[0] = 1;
    REP(i,N+2)tpow[i+1] = 2*tpow[i];
    ll answer = 0;
    for (ll i = N; i >= 0; i--){
        if(tpow[i+1]-2 >= X){
            X--;
            if(X < 0)X = 0;
        }else if(tpow[i+1] - 1 <= X){
            answer += tpow[i];
            X-= tpow[i+1] - 1;
        }
    }
    cout << answer << "\n";
}