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
#include<bits/stdc++.h>
#define ll long long

ll n,x,b[50],p[50];

ll c(ll n,ll x){
    if(n==0)return 1;
    if(x==0||x==1)return 0;
    else if(x<=1+b[n-1])return c(n-1,x-1);
    else if(x==2+b[n-1])return 1+p[n-1];
    else if(x<=2+2*b[n-1])return 1+p[n-1]+c(n-1,x-2-b[n-1]);
    else if(x==3+2*b[n-1])return p[n-1]*2LL+1;
}

int main(){
    scanf("%lld %lld",&n,&x);
    b[0]=1;
    p[0]=1;
    for(int i=1;i<50;i++)p[i]=2*p[i-1]+1,b[i]=2*b[i-1]+3;
    printf("%lld\n",c(n,x));
}