POJ2115 C Looooops

/*
 POJ2115 C Looooops
 http://poj.org/problem?id=2115
 扩展欧几里得
 题意：求x, s.t. (a+c*x)=b (mod 1<<k)
        <=> c*x=b-a (mod 1<<k)
 */
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <iostream>
#include <map>
#include <set>
//#define test
using namespace std;
const int Nmax=1e6+7;
long long a,b,c,k,x,y;
long long ex_gcd(long long a,long long b,long long &x,long long &y)//solve x,y in a*x+b*y=ex_gcd(a,b,x,y)=gcd(a,b);
{
    if(b==0)
    {
        x=1LL;
        y=0LL;
        return a;
    }
    long long ans=ex_gcd(b,a%b,x,y);
    long long tmp=x;
    x=y;
    y=tmp-a/b*y;
    return ans;
    //x = x0 + (b/gcd)*t
    //y = y0 – (a/gcd)*t
     
}
long long get(long long a,long long m,long long c)//get x in a*x=c(mod m)
{
    //we can solve x,y in a*x+b*y=c <=> c%gcd(a,b)==0
    long long x,y;
    long long gcd=ex_gcd(a,m,x,y);
    if(c%gcd!=0)
        return -1;//error
    m/=gcd;
    x*=c/gcd;
    if(m<0)
        m=-m;
    long long ans=x%m;
    while(ans<0)
        ans+=m;
    return ans;
}


int main()
{
    #ifdef test
    #endif
    //freopen("poj2115.in","r",stdin);
    while(1)
    {
        scanf("%lld%lld%lld%lld",&a,&b,&c,&k);
        if(!a&&!b&&!c&&!k)
            break;
        long long d=b-a;
        //if(d<0)//即使是负数也不用考虑，可以得出正确答案
            //d+=(1LL<<k);
        long long ans=get(c,(1LL<<k),d);
        if(ans==-1)
            printf("FOREVER
");
        else
            printf("%lld
",ans);
    }
    return 0;
}