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 <iostream>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <iomanip>
#include <cmath>
typedef long long ll;
using namespace std;

void solve(){
  ll n, x, l = 0, r, ans = 0; cin >> n >> x;
  vector<ll> y(n + 1), sum(n + 1);
  y[0] = 1;
  sum[0] = 1;
  for(int i = 1; i <= n; i++){
    y[i] = y[i - 1] * 2 + 3;
    sum[i] = sum[i - 1] * 2 + 1;
  }
  r = y[n] - 1;
  for(int i = n; i >= 1; i--){
    ll mid = (l + r) / 2;
    if(x == mid){
      ans += sum[i] / 2 + 1;
      break;
    }
    if(x < mid){
      l++;
      r = mid - 1;
    }
    else{
      ans += sum[i] / 2 + 1;
      l = mid + 1;
      r--;
    }
  }
  cout << ans << endl;
  return;
}

int main(){
  solve();
  return 0;
}