Finding a House

Sanghyuk lives in Anyang and is worn out after four years of commuting, so he wants to find a house in Seoul. The house he wants meets all three conditions below.

• Near a McDonald's: the shortest distance from the house to the nearest McDonald's is at most $x$.
• Near a Starbucks: the shortest distance from the house to the nearest Starbucks is at most $y$.
• Among the houses that meet both conditions above, the sum of the two shortest distances is the smallest.

The commute has worn Sanghyuk down and he cannot decide on a house. Solve the problem for him. A map of the area he is moving to is given as a weighted graph, and the McDonald's and Starbucks locations are given as vertex numbers. Write a program that prints the sum of the two shortest distances for the house Sanghyuk wants. Every vertex that has neither a McDonald's nor a Starbucks has a house.

In the map above, a square is a vertex with a McDonald's, a star is a vertex with a Starbucks, and each circle is a vertex with a house. When $x$ is 6 and $y$ is 4, the house that answers the question is vertex 6. Its shortest distance to a McDonald's is 2 and its shortest distance to a Starbucks is 4, so the sum is 6. Vertex 7 also meets both conditions, but its two shortest distances are 6 and 2 for a sum of 8, which is larger than the value at vertex 6, so it is not the answer. Vertices 2, 3 and 4 fail to meet both conditions at once, so none of them can be the answer.

The first line has the number of vertices $V(3 \le V \le 10000)$ and the number of roads $E(0 \le E \le 300000)$. Each of the next $E$ lines has three integers $u$, $v$ and $w$ in that order. They mean there is a road of weight $w(1 \le w < 10000)$ between vertices $u$ and $v(1 \le u, v \le V)$. $u$ and $v$ are different, and several roads may join the same pair of vertices.

The next line has the number of McDonald's locations $M(1 \le M \le V-2)$ and the threshold $x(1 \le x \le 100000000)$, and the line after it has the $M$ vertex numbers that have a McDonald's. The next line has the number of Starbucks locations $S(1 \le S \le V-2)$ and the threshold $y(1 \le y \le 100000000)$, and the line after it has the $S$ vertex numbers that have a Starbucks.

• A vertex with a McDonald's or a Starbucks has no house.
• One vertex may have both a McDonald's and a Starbucks.
• At least one vertex has a house.

Print the sum of the shortest distance to a McDonald's and the shortest distance to a Starbucks for the house Sanghyuk wants. If no house meets the conditions, print -1.