#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const int mod = 21252;

ll a[4],m[4];

ll M;

ll ex_gcd(ll a,ll b,ll &x,ll &y)

{

    if(b==0){x=1,y=0;return a;}

    ll d=ex_gcd(b,a%b,x,y);

    ll temp=x;x=y;y=temp-a/b*y;

    return d;

}

ll crt(ll n)

{

    ll i,Mi,xi,yi,d,Ans=0;

    M=1;

    for(i=1;i<=n;i++)M*=m[i];

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

    {

        Mi=M/m[i];d=ex_gcd(Mi,m[i],xi,yi);

        Ans=(Ans+Mi*a[i]*xi)%M;

    }

    return (Ans%M+M)%M;//最小整数解，因为gcd为1所以直接M就可以了

}

int main()

{

    ll k,t=0;

    m[1]=23,m[2]=28,m[3]=33;

    while(~scanf("%lld%lld%lld%lld",&a[1],&a[2],&a[3],&k),a[1]!=-1||a[2]!=-1||a[3]!=-1||k!=-1)

    {

        ll ans=crt(3);

        while(ans<=k) ans+=M;//要超过k才行

        printf("Case %lld: the next triple peak occurs in ",++t);

        cout<<ans-k<<" days.\n";

    }

    return 0;

 }