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#include <cstdio>
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#include <iostream>
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#include <stdlib.h>
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using namespace std;
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#define int long long
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#define endl "\r"
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const int N = 10;
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int n = 3;
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int a[N] = {0, 23, 28, 33};
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int m[N];
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int exgcd(int a, int b, int &x, int &y) {
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if (!b) {
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x = 1, y = 0;
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return a;
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}
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int d = exgcd(b, a % b, y, x);
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y -= a / b * x;
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return d;
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}
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int exCRT() { // m代表合并多少次，有n个方程，其实m=n-1
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int a1 = a[1], m1 = m[1], a2, m2, k1, k2;
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for (int i = 2; i <= n; i++) { // a1,m1：做为先遣军，准备与后续的方程合并
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a2 = a[i], m2 = m[i];
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int d = exgcd(a1, -a2, k1, k2); // 扩展欧几里得求方程特解,x有用，y无用
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if ((m2 - m1) % d) return -1; // 拼出来的同余方程无解，继续不下去
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k1 *= (m2 - m1) / d; // 同比扩大指定倍数
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int t = abs(a2 / d); // 最小累计单元
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k1 = (k1 % t + t) % t; // 取得最小正整数解，否则可以会在计算过程中爆long long
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m1 = k1 * a1 + m1; // 准备下一轮的m[1],a[1]
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a1 = abs(a1 / d * a2); // 两个顺序不能反
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}
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return (m1 % a1 + a1) % a1; // 取得最小正整数解
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}
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int T;
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int m1, m2, m3, d;
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signed main() {
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while (cin >> m[1] >> m[2] >> m[3] >> d && ~m[1]) {
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T++;
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int res = exCRT();
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int A = a[1] * a[2] * a[3];
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if (res <= d) res += A;
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printf("Case %lld: the next triple peak occurs in %lld days.\n", T, res - d);
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}
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}
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