One-Way Reform

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
分析：对于度数为奇数的节点，必然存在一条子欧拉路径，那么除去两个端点，其他的都是对答案有贡献的；
　　　然后对于子欧拉回路，每个节点都有贡献；
　　　为了防止重复计数，vis数组判断有没有访问过；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
const int maxn=2e2+;
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
inline ll read()
{
    ll x=;int f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
int n,m,k,t,now[maxn],du[maxn],cnt,ret;
bool c[maxn][maxn],vis[maxn];
vector<pii>ans;
void dfs(int p)
{
    if(!vis[p])vis[p]=true,cnt++;
    for(;now[p]<=n;now[p]++)
    {
        int q=now[p];
        if(c[p][q])
        {
            ans.pb(mp(p,q));
            --du[p],--du[q];
            c[p][q]=c[q][p]=false;
            dfs(q);
            break;
        }
    }
}
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        ans.clear();
        ret=;
        memset(now,,sizeof(now));
        memset(vis,false,sizeof(vis));
        scanf("%d%d",&n,&m);
        while(m--)
        {
            scanf("%d%d",&j,&k);
            du[j]++,du[k]++;
            c[j][k]=c[k][j]=true;
        }
        for(i=;i<=n;i++)
        {
            if(du[i]&)
            {
                cnt=;
                dfs(i);
                ret+=cnt-;
            }
        }
        for(i=;i<=n;i++)
        {
            if(du[i])
            {
                 cnt=;
                 dfs(i);
                 ret+=cnt;
            }
            else if(!du[i]&&!vis[i])ret++;
        }
        printf("%d\n",ret);
        for(pii x:ans)printf("%d %d\n",x.fi,x.se);
    }
    //system("Pause");
    return ;
}