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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> iint;
typedef pair<ll,ll> llll;
#define ALL(x) (x).begin(),(x).end()
const ll zero = 0;
const ll one = 1;
const ll INF = 9223372036854775807; //10^18
const int inINF = 2147483647; //10^9
const ll MOD = 1000000007; //10^9+7
const ll MOD2 = 998244353;
void Yes() {printf("Yes\n");}
void No() {printf("No\n");}
void YES() {printf("YES\n");}
void NO() {printf("NO\n");}

vector<long long> a, p;

long long f(ll N, ll X){
    if(X == 0){
        return 0;
    }
    else if(N == 1){
        return 1;
    }
    else if(X <= a[N-1] + 1){
        return f(N-1, X-1);
    }
    else{
        return p[N-1] + 1 + f(N-1, X - a[N-1] - 2);
    }
}

int main(){
    a.resize(51); p.resize(51);
    a[0] = 1;
    p[0] = 1;
    for (ll i = 1; i < 51; i++) {
        a[i] = 2 * a[i-1] + 3;
        p[i] = 2 * p[i-1] + 1;
    }
    ll N, X;
    cin >> N >> X;
    printf("%lld\n", f(N, X));

}