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 Python (3.4.3) (judge verdict: Accepted)

# Python3 (3.4.3)
import sys
input = sys.stdin.readline

# -------------------------------------------------------------
# function
# -------------------------------------------------------------
def dfs(N,X):
    # Lv.0
    if N == 0:
        # 調べる数が残っていれば : 1
        if X > 0:
            return 1
        # 残ってなければ : 0
        else:
            return 0
    # Lv.1以上
    else:
        # 下側の Lv.n-1 までの場合
        if X <= 1 + T[N-1]:
            return dfs(N-1,X-1)
        # 真ん中のパティまでの場合
        elif X == 1 + T[N-1] + 1:
            return P[N-1] + 1
        # 上側の Lv.n-1 までの場合
        else:
            return P[N-1] + 1 + dfs(N-1,X-2-T[N-1])

# -------------------------------------------------------------
# main
# -------------------------------------------------------------
N,X = map(int,input().split())

# 事前計算
# T[i] : Lv i バーガーの層数
# P[i] : Lv i バーガーのパティ数
T,P = [0]*(N+1),[0]*(N+1)
T[0],P[0] = 1,1
for i in range(N):
    T[i+1] = T[i]*2 + 3
    P[i+1] = P[i]*2 + 1

print(dfs(N,X))