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;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef pair<ll, ll> pii;
typedef vector<pii> vpii;

ll INF = (1ll << 32);
double EPS = 1e-4;

ll MOD = 1000000007;

namespace std {
  template <>
  class hash<pii> {
  public:
    size_t operator()(const pii& x) const {
      return hash<ll>()(x.first) ^ hash<ll>()(x.second);
    }
  };
}

unordered_map<pii, ll> cache;

ll f(ll a, ll b) {
  if (b == 0) return 0;
  if (a == 0) return 1;

  if (cache.find(make_pair(a, b)) != cache.end()) {
    return cache[make_pair(a, b)];
  }

  ll h = 1;
  for (ll i = 0; i < a; i++) {
    h = h * 2 + 3;
  }

  ll prev_h = (h - 3) / 2;

  ll ret = 0;

  if (b >= 2) ret += f(a - 1, min(prev_h, b - 1));
  if (b >= 2 + prev_h) ret += 1;
  if (b >= 3 + prev_h) ret += f(a - 1, min(prev_h, b - 2 - prev_h));

  cache[make_pair(a, b)] = ret;

  return ret;
}

int main() {
  ll N, X;
  cin >> N >> X;

  cout << f(N, X) << endl;
}