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<cstdio>

int n;
long x, ans;
long po[51];

void dfs(int n, long x) {
    if (x == 2 * po[n] - 1) ans += po[n];
    else if (x >= po[n]) {
        ans += po[n - 1] + 1;
        dfs(n - 1, x - po[n]);
    }
    else if (x > 0) {
        dfs(n - 1, x - 1);
    }
}

int main() {
    scanf("%d %ld", &n, &x);
    po[0] = 1;
    for (int i = 1; i <= n; i++) po[i] = 2 * po[i - 1] + 1;
    dfs(n, x);
    printf("%ld\n", ans);
    return 0;
}   
