題目連結

E. One-Way Reform

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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

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.

Example

input

2
5 5
2 1
4 5
2 3
1 3
3 5
7 2
3 7
4 2

output

3
1 3
3 5
5 4
3 2
2 1
3
2 4
3 7

題意：

n個點m條邊的無向連通圖 
沒有自環沒有重邊 
我們要把所有點都定向 
希望使得儘可能多的點擁有相同的入度與出度 
讓你輸出滿足這個條件的最大點數和每條邊最後的定向 

題解：

首先計算所有點的度，如果存在度為奇數的點，那麼這種點的個數必定為偶數（因為由於每條邊在度數總和上增加2，那麼所以奇數度的點的度只和也必定為奇數），那麼我們新建一個結點n+1，將那些度為奇數的點連向n+1。那麼這個此時這個圖每個點度都為偶數了，必定有歐拉回路，使用Fleury演算法畫尤拉路就可以了。

另外要注意圖可能不連通,所以要對每個點進行dfs。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<set>
using namespace std;
const int MAXN=200+100;
set<int> g[MAXN];
int deg[MAXN];
vector<pair<int,int> >res;
int n,m;
void init()
{
	memset(deg,0,sizeof(deg));
	for(int i=0;i<MAXN;i++) g[i].clear();
	res.clear();
}
void dfs(int u)
{
	while(g[u].size())
	{
		int v=*g[u].begin();
		g[u].erase(v),g[v].erase(u);
		if(u!=n+1&&v!=n+1)
		 printf("%d %d\n",u,v);
		dfs(v);
	}
}
int main()
{
	int cas;
	scanf("%d",&cas);
	while(cas--)
	{
		init();
		scanf("%d%d",&n,&m);
		while(m--)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			g[u].insert(v),g[v].insert(u);
			deg[u]++,deg[v]++;
		}
		int ans=0;
		for(int i=1;i<=n;i++)
		if(deg[i]&1) ans++,g[i].insert(n+1),g[n+1].insert(i);
		printf("%d\n",n-ans);
		for(int i=1;i<=n;i++) dfs(i);
		
	}
	return 0;
}