"Oh, There is a bipartite graph.""Make it Fantastic."

X wants to check whether a bipartite graph is a fantastic graph. He has two fantastic numbers, and he wants to let all the degrees to between the two boundaries. You can pick up several edges from the current graph and try to make the degrees of every point to between the two boundaries. If you pick one edge, the degrees of two end points will both increase by one. Can you help X to check whether it is possible to fix the graph?

Input

There are at most 3030 test cases.

For each test case,The first line contains three integers \$N\$ the number of left part graph vertices, \$M\$ the number of right part graph vertices, and \$K\$ the number of edges (\$1 \le N \le 2000,0 \le M \le 2000,0 \le K \le 6000\$). Vertices are numbered from \$1\$ to \$N\$.

The second line contains two numbers L, RL,R (\$0 \le L \le R \le 300\$). The two fantastic numbers.

Then \$K\$ lines follows, each line containing two numbers \$U, V (1 \le U \le N,1 \le V \le M)\$. It shows that there is a directed edge from \$U\$-th spot to \$V\$-th spot.

Note. There may be multiple edges between two vertices.

Output

One line containing a sentence. Begin with the case number. If it is possible to pick some edges to make the graph fantastic, output "Yes" (without quote), else output "No" (without quote).

```3 3 7
2 3
1 2
2 3
1 3
3 2
3 3
2 1
2 1
3 3 7
3 4
1 2
2 3
1 3
3 2
3 3
2 1
2 1```

```Case 1: Yes
Case 2: No```

\$s\$ 对于左侧的 \$N\$个点都连边，流量为 \$[L,R]\$；右侧的 \$M\$ 个点对 \$t\$ 都连边，流量为 \$[L,R]\$。

AC代码：

```#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=;
const int maxk=;

struct Edge{
int u,v,c,f;
};
struct Dinic
{
int s,t; //源点汇点
vector<Edge> E;
vector<int> G[maxn];
void init(int l,int r)
{
E.clear();
for(int i=l;i<=r;i++) G[i].clear();
}
void addedge(int from,int to,int cap)
{
E.push_back((Edge){from,to,cap,});
E.push_back((Edge){to,from,,});
G[from].push_back(E.size()-);
G[to].push_back(E.size()-);
}
int dist[maxn],vis[maxn];
queue<int> q;
bool bfs() //在残量网络上构造分层图
{
memset(vis,,sizeof(vis));
while(!q.empty()) q.pop();
q.push(s);
dist[s]=;
vis[s]=;
while(!q.empty())
{
int now=q.front(); q.pop();
for(int i=;i<G[now].size();i++)
{
Edge& e=E[G[now][i]]; int nxt=e.v;
if(!vis[nxt] && e.c>e.f)
{
dist[nxt]=dist[now]+;
q.push(nxt);
vis[nxt]=;
}
}
}
return vis[t];
}
int dfs(int now,int flow)
{
if(now==t || flow==) return flow;
int rest=flow,k;
for(int i=;rest> && i<G[now].size();i++)
{
Edge &e=E[G[now][i]]; int nxt=e.v;
if(e.c>e.f && dist[nxt]==dist[now]+)
{
k=dfs(nxt,min(rest,e.c-e.f));
if(!k) dist[nxt]=; //剪枝，去掉增广完毕的点
e.f+=k; E[G[now][i]^].f-=k;
rest-=k;
}
}
return flow-rest;
}
int mf; //存储最大流
int maxflow()
{
mf=;
int flow=;
while(bfs()) while(flow=dfs(s,INF)) mf+=flow;
return mf;
}
}dc;

int n,m,k;
int L,R;
int M[maxn];
int main()
{
int kase=;
while(cin>>n>>m>>k)
{

cin>>L>>R;

dc.init(,n+m+);
dc.s=n+m+;
dc.t=n+m+;

int S=,T=n+m+;
for(int u,v,i=;i<=k;i++)
{
scanf("%d%d",&u,&v);
}
memset(M,,sizeof(M));
for(int i=;i<=n;i++)
{
M[i]+=L;
M[S]-=L;
}
for(int i=;i<=m;i++)
{
M[T]+=L;
M[n+i]-=L;
}

int tmp=;
for(int i=;i<=n+m+;i++)
{
}

printf("Case %d: ",++kase);
if(dc.maxflow()==tmp) printf("Yes\n");
else printf("No\n");
}
}```

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