#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n, m, k, l, r, s, t, superS, superT;
const int MAXN = 5000;//点数的最大值
const int MAXM = 120000;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge
{int to,next,cap,flow;} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
void init()
{tol = 2;memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int rw = 0)
{edge[tol].to = v;edge[tol].cap = w;edge[tol].flow = 0;edge[tol].next = head[u];head[u] = tol++;edge[tol].to = u;edge[tol].cap = rw;edge[tol].flow = 0;edge[tol].next = head[v];head[v] = tol++;
}
int Q[MAXN];
int dep[MAXN],cur[MAXN],sta[MAXN];
bool bfs(int s,int t,int n)
{int front = 0,tail = 0;memset(dep,-1,sizeof(dep[0])*(n+1));dep[s] = 0;Q[tail++] = s;while(front < tail){int u = Q[front++];for(int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].to;if(edge[i].cap > edge[i].flow && dep[v] == -1){dep[v] = dep[u] + 1;if(v == t)return true;Q[tail++] = v;}}}return false;
}
int dinic(int s,int t,int n)
{int maxflow = 0;while(bfs(s,t,n)){for(int i = 0; i < n; i++)cur[i] = head[i];int u = s, tail = 0;while(cur[s] != -1){if(u == t){int tp = INF;for(int i = tail-1; i >= 0; i--)tp = min(tp,edge[sta[i]].cap-edge[sta[i]].flow);maxflow += tp;for(int i = tail-1; i >= 0; i--){edge[sta[i]].flow += tp;edge[sta[i]^1].flow -= tp;if(edge[sta[i]].cap-edge[sta[i]].flow == 0)tail = i;}u = edge[sta[tail]^1].to;}else if(cur[u] != -1 && edge[cur[u]].cap > edge[cur[u]].flow && dep[u] + 1 == dep[edge[cur[u]].to]){sta[tail++] = cur[u];u = edge[cur[u]].to;}else{while(u != s && cur[u] == -1)u = edge[sta[--tail]^1].to;cur[u] = edge[cur[u]].next;}}}return maxflow;
}
int main()
{int kase = 1;while(~scanf("%d %d %d", &m, &n, &k)){scanf("%d %d", &l,  &r);s = n + m + 1;t = n + m + 2;superS = n + m + 3;superT = n + m + 4;init();for(int i = 1; i <= k; i++){int u, v;scanf("%d %d", &u , &v);addedge(u, v + n, 1); //二分图 建一条容量为1的边
        }addedge(t,s,INF); //从汇点向源点建一条inf的边for(int i = 1; i <= n; i++){addedge(s, i, r - l); //从源点向左半图连边
            addedge(superS, i, l);addedge(s, superT, l);}for(int i = 1; i <= m; i++){int aim = i + n;addedge(aim, t, r - l);addedge(aim, superT, l);addedge(superS, t, l);}printf("Case %d: ", kase++);int ans = dinic(superS, superT, n+m+4);if(ans == (n+m)*l) puts("Yes");else puts("No");}return 0;
}