//求咬尾数，一个2^n位环形二进制数，该二进制的每n位连续二进制数都不同

//DFS求解欧拉回路

//Time:32ms     Memory:1668K

#include<iostream>

#include<cstring>

#include<cstdio>

using namespace std;

#define MAX 16

#define MAXK (1<<15)

int circle[MAX][MAXK];

bool v[MAXK];

int n;

bool dfs(int rk, int x)

{

    int t[2], mod = 1 << n;

    if(rk == mod + n - 1)   return true;    //可生成所有n位二进制数

    t[0] = (x << 1) & (mod - 1);    //向后移动补0

    t[1] = t[0] | 1;    //向后移动补1

    for(int i = 0; i < 2; i++)

    {

        if(!v[t[i]]){

            v[t[i]] = true;

            if(rk >= mod)  //最后一位超过该咬尾数，可成环继续，否则回溯

            {

                if(circle[n][rk%mod] != (t[i] & 1))

                    return false;

            }

            else circle[n][rk] = i;

            if(dfs(rk+1, t[i])) return true;

            v[t[i]] = false;

        }

    }

    return false;

}

int main()

{

    //freopen("in.txt","r",stdin);

    for(n = 1; n < MAX; n++)

    {

        memset(v,false,sizeof(v));

        v[0] = true;

        dfs(n,0);

    }

    int k;

    while(scanf("%d%d", &n,&k), n)

    {

        int sum = 0, mod = 1 << n;

        for(int i = 0; i < n; i++)

        {

            sum += circle[n][(i+k)%mod]*(1 << (n - i - 1));

        }

        printf("%d\n",sum);

    }

    return 0;

}