1067

/*
	威佐夫博奕（Wythoff Game） 两堆
	对于这种博弈，有一个通项公式：a[k] = [k*(1+sqrt(5))/2],b[k] = a[k]+k;

	基于两个规律.首先枚举前几个必败态
	(0,0) (1,2) (3,5) (4,7) (6,10),....
	规律是，a[k] = 等于之前没有出现的自然数
			b[k] = a[k]+k
	
	通项公式的给出，还有一个定理，beatty定理：
	正无理数，alpha,beta，如果有1/a+1/b=1,那么数列[an],[bn],严格递增，而且构成正整数上的一个划分

	a[n] = [cn]
	b[n] = a[n]+n = [c*n] +n = [(c+1)n] 
	所以有 1/c+1/(c+1) = 1
	解得，c=(1+sqrt(5))/2;
	完毕
*/
// include file
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <ctime>

#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <bitset>

#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <list>
#include <functional>

using namespace std;

// typedef
typedef __int64 LL;

// 
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)

#define Z(a,b) ((a)<<(b))
#define Y(a,b) ((a)>>(b))

const double eps = 1e-6;
const double INFf = 1e100;
const int INFi = 1000000000;
const LL INFll = (LL)1<<62;
const double Pi = acos(-1.0);

template<class T> inline T sqr(T a){return a*a;}
template<class T> inline T TMAX(T x,T y)
{
	if(x>y) return x;
	return y;
}
template<class T> inline T TMIN(T x,T y)
{
	if(x<y) return x;
	return y;
}
template<class T> inline T MMAX(T x,T y,T z)
{
	return TMAX(TMAX(x,y),z);
}
template<class T> inline T MMIN(T x,T y,T z)
{
	return TMIN(TMIN(x,y),z);
}
template<class T> inline void SWAP(T &x,T &y)
{
	T t = x;
	x = y;
	y = t;
}


// code begin
int a,b;
int main()
{
	read;
	write;
	int k,aa,bb;
	while(scanf("%d %d",&a,&b)==2)
	{
		if(a>b) SWAP(a,b);
		k = (int)(a*(sqrt(5.0)-1)/2);
		
		aa = (int)(k*(1+sqrt(5.0))/2);
		bb = aa+k;
		if(aa==a && bb==b)
		{
			printf("0\n");
			continue;
		}
		aa = (int)((k+1)*(1+sqrt(5.0))/2);
		bb = aa+k+1;
		if(aa==a && bb==b)
		{
			printf("0\n");
			continue;
		}
		printf("1\n");
	}
	return 0;
}