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.

Examples

Input

2 7


Output

4


Input

1 1


Output

0


Input

50 4321098765432109


Output

2160549382716056


import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		long x = sc.nextLong();

		String burger = "P";

		len = new long[51];
		len[0] = 1;

		numOfPate = new long[51];
		numOfPate[0] = 1;
		for(int i = 1; i <= 50; i++){
			len[i] = 2*len[i - 1] + 3;
			numOfPate[i] = 2*numOfPate[i - 1] + 1;
		}

//		for(int i = 1; i <= 5; i++){
//			System.out.println("Lebel "+i + " length "+ len[i]+ " pate "+numOfPate[i]);
//
//			burger = "B" + burger + "P" + burger + "B";
//
//			System.out.println(burger);
//		}

		System.out.println(solve(n, x));
	}

	static long len[];
	static long numOfPate[];

	//endを食べた層の直後のindexとする
	static long solve(int level, long end){
		long mid = len[level]/2;

		if(end <= level ){
			return 0;
		}

		if(end >= len[level] - level){
			return numOfPate[level];
		}

		//左半分の区間
		long result = solve(level - 1, end - 1);

		//センター分加算
		if(end > mid){
			result++;

			//右半分加算
			result += solve(level - 1, end - mid - 1);
		}

		return result;
	}

}
