#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cassert>
using namespace std;
int a, b, maxd;
typedef long long LL;
LL gcd(LL a, LL b)  //辗转相除法，求最大公因子
{return b == 0 ? a : gcd(b, a%b);
}
// 返回满足1/c <= a/b的最小c
inline int get_first(LL a, LL b)
{return b/a+1;
}
const int maxn = 100 + 5;
LL v[maxn], ans[maxn];
// 如果当前解v比目前最优解ans更优，更新ans这一判断的依据
bool better(int d)  //前d个，这时d只能是maxd，
{for(int i = d; i >= 0; i--) if(v[i] != ans[i]){return ans[i] == -1 || v[i] < ans[i];//没被访问过，或者是当前序列更优}return false;
}
// 当前深度为d，分母(注意是分母)不能小于from，分数之和恰好为aa/bb
bool dfs(int d, int from, LL aa, LL bb)
{if(d == maxd){if(bb % aa) return false; // aa/bb必须是埃及分数（任何数mod1都等于0，可以起到一部分剪枝的作用）v[d] = bb/aa;//不用再进一步展开分数if(better(d)) memcpy(ans, v, sizeof(LL)*(d+1));return true;}bool ok = false;//最后一层没找到解，可向上返回from = max(from, get_first(aa, bb)); // 枚举的起点( 越大，取倒之后就越小）for(int i = from; ; i++)  //进行枚举{// 剪枝：如果剩下的maxd+1-d个分数全部都是1/i，加起来仍然不超过aa/bb，则无解if(bb * (maxd+1-d) <= i * aa) break;//从第0层开始算的，总共有maxn+1个v[d] = i;// 计算aa/bb - 1/i，设结果为a2/b2LL b2 = bb*i;LL a2 = aa*i - bb;LL g = gcd(a2, b2); // 以便约分if(dfs(d+1, get_first(a2/g, b2/g), a2/g, b2/g)) ok = true;}return ok;//注意这些返回条件
}int main()
{int kase = 0;while(cin >> a >> b){int ok = 0;for(maxd = 1; maxd <= 100; maxd++)  //从小到大，枚举深度上限{memset(ans, -1, sizeof(ans));if(dfs(0, get_first(a, b ), a, b)){ok = 1;//找到解了~~break;}}cout << "Case " << ++kase << ": ";if(ok){cout << a << "/" << b << "=";for(int i = 0; i < maxd; i++) cout << "1/" << ans[i] << "+";cout << "1/" << ans[maxd] << "\n";}else cout << "No solution.\n";}return 0;
}