#include <bits/stdc++.h>
using namespace std;const int maxn = 52 + 10;
const long long INF = 1LL << 60;int n;
int G[maxn][maxn];
bool mark[maxn];
int p;
int src;
int dest;
long long d[maxn];int read_node()
{char ch[9];scanf("%s", ch);if(ch[0] >= 'A' && ch[0] <= 'Z') return ch[0] - 'A';else return ch[0] - 'a' + 26;
}char format_node(int u)
{return u < 26 ? 'A' + u : 'a' + (u - 26);
}// 拿着item个东西去结点next，还剩多少个东西
long long forward(long long item, int next)
{if(next < 26) return item - (item + 19) / 20;return item - 1;
}// 至少要拿着多少个东西到达结点u，交税以后还能剩d[u]个东西
long long back(int u)
{if(u >= 26) return d[u]+1;long long X = d[u] * 20 / 19; // 初始值while(forward(X, u) < d[u]) X++; // 调整return X;
}void solve()
{n = 52; // 总是有52个结点memset(mark, 0, sizeof(mark));d[dest] = p;mark[dest] = 1;for(int i = 0; i < n; i++) if(i != dest){d[i] = INF;if(G[i][dest]) d[i] = back(dest);}// Dijkstra主过程，逆推while(!mark[src]){// 找最小的dint minu = -1;for(int i = 0; i < n; i++) if(!mark[i]){if(minu < 0 || d[i] < d[minu]) minu = i;}mark[minu] = 1;// 更新其他结点的dfor(int i = 0; i < n; i++) if(!mark[i]){if(G[i][minu]) d[i] = min(d[i], back(minu));}}printf("%lld\n", d[src]);printf("%c", format_node(src));int u = src;long long item = d[src];while(u != dest){int next;for(next = 0; next < n; next++) // 找到第一个可以走的结点if(G[u][next] && forward(item, next) >= d[next]) break;item = d[next];printf("-%c", format_node(next));u = next;}printf("\n");
}int main()
{int kase = 0;while(scanf("%d", &n) == 1 && n >= 0){memset(G, 0, sizeof(G));for(int i = 0; i < n; i++){int u = read_node();int v = read_node();if(u != v) G[u][v] = G[v][u] = 1;}scanf("%d", &p);src = read_node();dest = read_node();printf("Case %d:\n", ++kase);solve();}return 0;
}