1184_B2. The Doctor Meets Vader (Medium)
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.
```python

def naiveSolve():
    
    
    
    return

def floyd_warshall(n, edges):
    dist = [[0 if i == j else float("inf") for i in range(n)] for j in range(n)]
    # dist = [[float("inf") for i in range(n)] for j in range(n)] # if nodes are not directly connected to themselves
    # pred = [[None] * n for _ in range(n)]

    for u, v, d in edges:
        dist[u][v] = d
        # pred[u][v] = u

    for k in range(n):
        for i in range(n):
            for j in range(n):
                if dist[i][k] + dist[k][j] < dist[i][j]:
                    dist[i][j] = dist[i][k] + dist[k][j]
                    # pred[i][j] = pred[k][j]
    return dist
    # return dist, pred

def hopcroft_karp(graph, n, m):
    """
    Maximum bipartite matching using Hopcroft-Karp algorithm, running in O(|E| sqrt(|V|))
    """
    assert (n == len(graph))
    match1 = [-1] * n
    match2 = [-1] * m

    # Find a greedy match for possible speed up
    for node in range(n):
        for nei in graph[node]:
            if match2[nei] == -1:
                match1[node] = nei
                match2[nei] = node
                break
    while 1:
        bfs = [node for node in range(n) if match1[node] == -1]
        depth = [-1] * n
        for node in bfs:
            depth[node] = 0

        for node in bfs:
            for nei in graph[node]:
                next_node = match2[nei]
                if next_node == -1:
                    break
                if depth[next_node] == -1:
                    depth[next_node] = depth[node] + 1
                    bfs.append(next_node)
            else:
                continue
            break
        else:
            break

        pointer = [len(c) for c in graph]
        dfs = [node for node in range(n) if depth[node] == 0]
        while dfs:
            node = dfs[-1]
            while pointer[node]:
                pointer[node] -= 1
                nei = graph[node][pointer[node]]
                next_node = match2[nei]
                if next_node == -1:
                    # Augmenting path found
                    while nei != -1:
                        node = dfs.pop()
                        match2[nei], match1[node], nei = node, nei, match1[node]
                    break
                elif depth[node] + 1 == depth[next_node]:
                    dfs.append(next_node)
                    break
            else:
                dfs.pop()
    return match1, match2

def main():
    
    n,m=readIntArr()
    
    adj=[set() for _ in range(n+1)]
    for _ in range(m):
        u,v=readIntArr()
        adj[u].add(v)
        adj[v].add(u)
    
    s,b,k,h=readIntArr()
    # ships
    ships=[] # (location,attack,fuel)
    for _ in range(s):
        loc,atk,fuel=readIntArr()
        ships.append((loc,atk,fuel))
    bases=[] # (location,def)
    for _ in range(b):
        loc,deff=readIntArr()
        bases.append((loc,deff))
    
    edges=[]
    for u in range(1,n+1):
        for v in adj[u]:
            if u!=v:
                edges.append((u,v,1))
    dist=floyd_warshall(n+1,edges)
    
    graph=[[] for _ in range(s)] # graph[ship]=[base1,base2,...]
    for ship in range(s):
        for base in range(b):
            ls,atk,fuel=ships[ship]
            lb,deff=bases[base]
            if atk>=deff and fuel>=dist[ls][lb]:
                graph[ship].append(base)
    match1,match2=hopcroft_karp(graph,s,b)
    matchCnts=0
    for v in match1:
        if v!=-1:
            matchCnts+=1
    
    zeroDummiesCost=k*matchCnts
    # assuming all ships can reach a base, we try to block all ships
    shipsDummiesCost=h*s
    
    print(min(zeroDummiesCost,shipsDummiesCost))
    
    return



import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
# input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.

def oneLineArrayPrint(arr):
    print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
    print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
    print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
 
def readIntArr():
    return [int(x) for x in input().split()]
# def readFloatArr():
#     return [float(x) for x in input().split()]
 
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
    dv=defaultValFactory;da=dimensionArr
    if len(da)==1:return [dv() for _ in range(da[0])]
    else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
 
def queryInteractive(r):
    print('? {}'.format(r))
    sys.stdout.flush()
    return readIntArr()
 
def answerInteractive(adj,n):
    print('!')
    for u in range(1,n+1):
        for v in adj[u]:
            if v>u:
                print('{} {}'.format(u,v))
    sys.stdout.flush()
 
inf=float('inf')
MOD=10**9+7
# MOD=998244353

from math import gcd,floor,ceil
# from math import floor,ceil # for Python2
 
for _abc in range(1):
    main()
```