Problem: Thanks to the Doctor's help, the rebels managed to steal enough gold to launch a full-scale attack on the Empire! However, Darth Vader is looking for revenge and wants to take back his gold.

The rebels have hidden the gold in various bases throughout the galaxy. Darth Vader and the Empire are looking to send out their spaceships to attack these bases.

The galaxy can be represented as an undirected graph with n planets (nodes) and m wormholes (edges), each connecting two planets.

A total of s empire spaceships and b rebel bases are located at different planets in the galaxy.

Each spaceship is given a location x, denoting the index of the planet on which it is located, an attacking strength a, and a certain amount of fuel f.

Each base is given a location x, and a defensive strength d.

A spaceship can attack a base if both of these conditions hold: 

  * the spaceship's attacking strength is greater or equal than the defensive strength of the base 
  * the spaceship's fuel is greater or equal to the shortest distance, computed as the number of wormholes, between the spaceship's planet and the base's planet 



Vader is very particular about his attacking formations. He requires that each spaceship is to attack at most one base and that each base is to be attacked by at most one spaceship.

Vader knows that the rebels have hidden k gold in each base, so he will assign the spaceships to attack bases in such a way that maximizes the number of bases attacked.

Therefore, for each base that is attacked, the rebels lose k gold.

However, the rebels have the ability to create any number of dummy bases. With the Doctor's help, these bases would exist beyond space and time, so all spaceship can reach them and attack them. Moreover, a dummy base is designed to seem irresistible: that is, it will always be attacked by some spaceship.

Of course, dummy bases do not contain any gold, but creating such a dummy base costs h gold.

What is the minimum gold the rebels can lose if they create an optimal number of dummy bases?

Input

The first line contains two integers n and m (1 ≤ n ≤ 100, 0 ≤ m ≤ 10000), the number of nodes and the number of edges, respectively.

The next m lines contain two integers u and v (1 ≤ u, v ≤ n) denoting an undirected edge between the two nodes.

The next line contains four integers s, b, k and h (1 ≤ s, b ≤ 1000, 0 ≤ k, h ≤ 10^9), the number of spaceships, the number of bases, the cost of having a base attacked, and the cost of creating a dummy base, respectively.

The next s lines contain three integers x, a, f (1 ≤ x ≤ n, 0 ≤ a, f ≤ 10^9), denoting the location, attack, and fuel of the spaceship.

The next b lines contain two integers x, d (1 ≤ x ≤ n, 0 ≤ d ≤ 10^9), denoting the location and defence of the base.

Output

Print a single integer, the minimum cost in terms of gold.

Example

Input


6 7
1 2
2 3
3 4
4 6
6 5
4 4
3 6
4 2 7 3
1 10 2
3 8 2
5 1 0
6 5 4
3 7
5 2


Output


12

Note

One way to minimize the cost is to build 4 dummy bases, for a total cost of 4 × 3 = 12.

One empire spaceship will be assigned to attack each of these dummy bases, resulting in zero actual bases attacked.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;


public class  Main{
	static class base implements Comparable<base>{
		int d,g;
		base(int dd,int gg){
			d=dd;g=gg;
		}
		@Override
		public int compareTo(base o) {
			return d-o.d;
		}
		
	}
	static long inf=(long)1e15+7;
	static long[][]adj;
	
	static int V, s, t;
	static long res[][];			//input
	static ArrayList<Integer>[] adjList;	//input
	static int[] ptr, dist;
	
	static long dinic()						//O(V^2E)
	{
		long mf = 0;
		while(bfs())
		{
			ptr = new int[V];
			long f;
			while((f = dfs(s, inf)) != 0)
				mf += f;
		}
		return mf;
	}
	
	
	static boolean bfs()
	{
		dist = new int[V];
		Arrays.fill(dist, -1);
		dist[s] = 0;
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(s);
		while(!q.isEmpty())
		{
			int u = q.remove();
			if(u == t)
				return true;
			for(int v: adjList[u])
				if(dist[v] == -1 && res[u][v] > 0)
				{
					dist[v] = dist[u] + 1;
					q.add(v);
				}
		}
		return false;
	}
	
	static long dfs(int u, long flow)
	{
		if(u == t)
			return flow;
		for(int i = ptr[u]; i < adjList[u].size(); i = ++ptr[u])
		{
			int v = adjList[u].get(i);
			if(dist[v] == dist[u] + 1 && res[u][v] > 0)
			{
				long f = dfs(v, Math.min(flow, res[u][v]));
				if(f > 0)
				{
					res[u][v] -= f;
					res[v][u] += f;
					return f;
				}
			}
		}
		return 0;
		
	}
	public static void main(String[] args) throws Exception{
		MScanner sc=new MScanner(System.in);
		PrintWriter pw=new PrintWriter(System.out);
		int n=sc.nextInt(),m=sc.nextInt();
		adj=new long[n][n];
		for(long []i:adj)Arrays.fill(i, inf);
		for(int i=0;i<m;i++) {
			int x=sc.nextInt()-1,y=sc.nextInt()-1;
			adj[x][y]=1;
			adj[y][x]=1;
		}
		for(int i=0;i<n;i++)adj[i][i]=0;
		for(int k=0;k<n;k++) {
			for(int i=0;i<n;i++) {
				for(int j=0;j<n;j++) {
					adj[i][j]=Math.min(adj[i][j], adj[i][k]+adj[k][j]);
				}
			}
		}
		int ss=sc.nextInt(),b=sc.nextInt(),k=sc.nextInt();
		int[][]spaceShips=new int[ss][5];
		for(int i=0;i<ss;i++) {
			for(int j=0;j<4;j++) {
				spaceShips[i][j]=sc.nextInt();
			}
			spaceShips[i][0]--;
			spaceShips[i][4]=i;
		}
		Arrays.sort(spaceShips,(x,y) -> x[1]-y[1]);
		
		TreeSet<base>[]planets=new TreeSet[n];
		for(int i=0;i<n;i++) {
			planets[i]=new TreeSet<base>();
		}
		for(int i=0;i<b;i++) {
			planets[sc.nextInt()-1].add(new base(sc.nextInt(), sc.nextInt()));
		}
		
		long[]maxProfit=new long[ss];
		long[]maxSoFar=new long[n];
		Arrays.fill(maxSoFar, -inf);
		for(int i=0;i<ss;i++) {
			int curAtt=spaceShips[i][1],curFuel=spaceShips[i][2],curPrice=spaceShips[i][3],curLoc=spaceShips[i][0];
			long profit=-inf;
			for(int p=0;p<n;p++) {
				long dist=adj[curLoc][p];
				if(dist>curFuel)continue;
				
				while(!planets[p].isEmpty() && curAtt>=planets[p].first().d) {
					maxSoFar[p]=Math.max(maxSoFar[p], planets[p].pollFirst().g);
				}
				profit=Math.max(profit, maxSoFar[p]);
				
			}
			maxProfit[spaceShips[i][4]]=Math.max(-inf, profit*1l-curPrice);
		}
		long ans=0;
		
		HashSet<Integer>dep=new HashSet<Integer>();
		int[][]depend=new int[k][2];
		for(int i=0;i<k;i++) {
			int x=sc.nextInt()-1,y=sc.nextInt()-1;
			dep.add(x);dep.add(y);
			depend[i][0]=x;depend[i][1]=y;
		}
		
		
		V=dep.size()+2;
		res=new long[V][V];
		adjList=new ArrayList[V];
		for(int i=0;i<V;i++)adjList[i]=new ArrayList<Integer>();
		
		t=V-1;
		HashMap<Integer, Integer>ids=new HashMap<Integer,Integer>();
		int id=1;
		
		for(int i=0;i<k;i++) {
			int x=depend[i][0],y=depend[i][1];
			int idx,idy;
			if(ids.containsKey(x)) {
				idx=ids.get(x);
			}
			else {
				idx=id++;
				ids.put(x, idx);
			}
			if(ids.containsKey(y)) {
				idy=ids.get(y);
			}
			else {
				idy=id++;
				ids.put(y, idy);
			}
			
			adjList[idx].add(idy);
			adjList[idy].add(idx);
			res[idx][idy]=inf;
			
			
			
		}
		
		for(int i=0;i<ss;i++) {
			if(dep.contains(i)) {
				int curID=ids.get(i);
				if(maxProfit[i]>=0) {
					adjList[s].add(curID);
					adjList[curID].add(s);
					res[s][curID]=maxProfit[i];
				}
				else {
					adjList[t].add(curID);
					adjList[curID].add(t);
					res[curID][t]=-maxProfit[i];
				}
			}
			ans+=Math.max(0, maxProfit[i]);
		}
		pw.println(ans-dinic());
		pw.flush();
	}
	static class MScanner {
		StringTokenizer st;
		BufferedReader br;
		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public MScanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int[] intArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public long[] longArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public int[] intSortedArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public long[] longSortedArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		static void shuffle(int[]in) {
			for(int i=0;i<in.length;i++) {
				int idx=(int)(Math.random()*in.length);
				int tmp=in[i];
				in[i]=in[idx];
				in[idx]=tmp;
			}
		}
		static void shuffle(long[]in) {
			for(int i=0;i<in.length;i++) {
				int idx=(int)(Math.random()*in.length);
				long tmp=in[i];
				in[i]=in[idx];
				in[idx]=tmp;
			}
		}
		public Integer[] IntegerArr(int n) throws IOException {
	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public Long[] LongArr(int n) throws IOException {
	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
	static void addX(int[]in,int x) {
		for(int i=0;i<in.length;i++)in[i]+=x;
	}
	static void addX(long[]in,int x) {
		for(int i=0;i<in.length;i++)in[i]+=x;
	}
}