OPTM - Optimal Marks

给你一个无向图G（V，E）。 每个顶点都有一个int范围内的整数的标记。 不同的顶点可能有相同的标记。 对于边（u，v），我们定义Cost（u，v）= mark [u] xor mark [v]。 现在我们知道某些节点的标记了。你需要确定其他节点的标记，以使边的总成本尽可能小。 如果有多解，请输出 $\sum mark_i$ 最小的方案。如果仍有多解，输出任意一个。

You are given an undirected graph G(V, E). Each vertex has a mark which is an integer from the range \[0..2 $ ^{31} $ – 1\]. Different vertexes may have the same mark. For an edge (u, v), we define Cost(u, v) = mark\[u\] xor mark\[v\]. Now we know the marks of some certain nodes. You have to determine the marks of other nodes so that the total cost of edges is as small as possible.

The first line of the input data contains integer **T** (1 ≤ **T** ≤ 10) - the number of testcases. Then the descriptions of T testcases follow. First line of each testcase contains 2 integers **N** and **M** (0 < **N** <= 500, 0 <= **M** <= 3000). **N** is the number of vertexes and **M** is the number of edges. Then **M** lines describing edges follow, each of them contains two integers u, v representing an edge connecting u and v. Then an integer **K**, representing the number of nodes whose mark is known. The next **K** lines contain 2 integers u and p each, meaning that node u has a mark p. It’s guaranteed that nodes won’t duplicate in this part.

For each testcase you should print **N** lines integer the output. The **K**th line contains an integer number representing the mark of node **K**. If there are several solutions, you have to output the one which minimize the sum of marks. If there are several solutions, just output any of them.

1
3 2
1 2
2 3
2
1 5
3 100

5
4
100