One-Way Reform
Contest: Codeforces Round 375 (Div. 2)
Difficulty rating: 2200
Tags: constructive algorithms, dfs and similar, flows, graphs, greedy

There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible that there is no way to get from one city to some other city using only these roads.

The road minister decided to make a reform in Berland and to orient all roads in the country, i.e. to make each road one-way. The minister wants to maximize the number of cities, for which the number of roads that begins in the city equals to the number of roads that ends in it.

Input format:
The first line contains a positive integer t (1 ≤ t ≤ 200) — the number of testsets in the input.

Each of the testsets is given in the following way. The first line contains two integers n and m (1 ≤ n ≤ 200, 0 ≤ m ≤ n·(n - 1) / 2) — the number of cities and the number of roads in Berland.

The next m lines contain the description of roads in Berland. Each line contains two integers u and v (1 ≤ u, v ≤ n) — the cities the corresponding road connects. It's guaranteed that there are no self-loops and multiple roads. It is possible that there is no way along roads between a pair of cities.

It is guaranteed that the total number of cities in all testset of input data doesn't exceed 200.

Pay attention that for hacks, you can only use tests consisting of one testset, so t should be equal to one.

Output format:
For each testset print the maximum number of such cities that the number of roads that begins in the city, is equal to the number of roads that ends in it.

In the next m lines print oriented roads. First print the number of the city where the road begins and then the number of the city where the road ends. If there are several answers, print any of them. It is allowed to print roads in each test in arbitrary order. Each road should be printed exactly once.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Example input:
2
5 5
2 1
4 5
2 3
1 3
3 5
7 2
3 7
4 2

Example output:
3
1 3
3 5
5 4
3 2
2 1
3
2 4
3 7

Editorial:
Let's solve this problem for each connected component separately.
At first we need to understand fact that in each connected component there are even number of vertices with odd degree. Let in the current connected component there are k vertices with odd degree and they have numbers o1, o2, ..., ok. Then we need to add in graph non-directional edges (o1, o2), (o3, o4), ... (ok - 1, ok). So in the current component now all vertices have even degree and there is exist Euler cycle, which we need to find and orientate all edges in the order of this cycle. It is easy to show that after that for all vertices which had even degree before we added the edges in-degree is equal to out-degree. In the other words, we show that the maximum number of vertices with equal in- and out-degrees in orientated graph equals to the number of vertices with even degree in non-orientated graph.
After we found Euler cycles for all connected components we need to print orientation of the edges and be careful and do not print edges which we added to the graph to connect vertices with odd degrees.