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>
using namespace std;
typedef long long LL;

int N;
LL X, D[60];

LL f(int lv, LL k){
    if (k <= 0) return 0;
    LL r = (D[lv] - 1) / 2;
    if (k >= D[lv]) return (1ll << (lv+1)) - 1;
    if (k <= r) return f(lv-1, k-1);
    if (k == r+1) return 1ll << lv;
    return (1ll << lv) + f(lv-1, k-r-1);
}

int main(){
    scanf("%d %lld", &N, &X);
    D[0] = 1;
    for (int i=1; i<=N; i++) D[i] = 2*D[i-1] + 3;
    printf("%lld\n", f(N, X));
    return 0;
}
