Description

Consider a town where all the streets are one-way and each street leads from one intersection to another. It is also known that starting from an intersection and walking through town's streets you can never reach the same intersection i.e. the town's streets form no cycles.

With these assumptions your task is to write a program that finds the minimum number of paratroopers that can descend on the town and visit all the intersections of this town in such a way that more than one paratrooper visits no intersection. Each paratrooper lands at an intersection and can visit other intersections following the town streets. There are no restrictions about the starting intersection for each paratrooper.

Input

Your program should read sets of data. The first line of the input file contains the number of the data sets. Each data set specifies the structure of a town and has the format:

no_of_intersections
no_of_streets
S1 E1
S2 E2
......
Sno_of_streets Eno_of_streets

The first line of each data set contains a positive integer no_of_intersections (greater than 0 and less or equal to 120), which is the number of intersections in the town. The second line contains a positive integer no_of_streets, which is the number of streets in the town. The next no_of_streets lines, one for each street in the town, are randomly ordered and represent the town's streets. The line corresponding to street k (k <= no_of_streets) consists of two positive integers, separated by one blank: Sk (1 <= Sk <= no_of_intersections) - the number of the intersection that is the start of the street, and Ek (1 <= Ek <= no_of_intersections) - the number of the intersection that is the end of the street. Intersections are represented by integers from 1 to no_of_intersections.

There are no blank lines between consecutive sets of data. Input data are correct.

Output

The result of the program is on standard output. For each input data set the program prints on a single line, starting from the beginning of the line, one integer: the minimum number of paratroopers required to visit all the intersections in the town.

Sample Input

2
4
3
3 4
1 3
2 3
3
3
1 3
1 2
2 3

Sample Output

2
1

rt

好惭愧到现在才会==

原理：

最小路径覆盖(path covering)：是“路径” 覆盖“点”，即用尽量少的不相交简单路径覆盖有向无环图G的所有顶点，即每个顶点严格属于一条路径。路径的长度可能为0(单个点)。

最小路径覆盖数＝G的点数－最小路径覆盖中的边数。应该使得最小路径覆盖中的边数尽量多，但是又不能让两条边在同一个顶点相交。拆点：将每一个顶点i拆成两个顶点Xi和Yi。然后根据原图中边的信息，从X部往Y部引边。所有边的方向都是由X部到Y部。因此，所转化出的二分图的最大匹配数则是原图G中最小路径覆盖上的边数。因此由最小路径覆盖数＝原图G的顶点数－二分图的最大匹配数便可以得解。（加单向边！！）

    #include <iostream>
    #include<cstdio>
    #include<cstring>
    #include<vector>
    using namespace std;
    #define M 550
    #define MAXN 240
    #define inf 0x3f3f3f3f
    int uN,vN;//u,v数目
    int g[MAXN][MAXN];
    int linker[MAXN];
    bool used[MAXN];
    bool dfs(int u)//从左边开始找增广路径
    {
        int v;
        for(v=0;v<vN;v++)//这个顶点编号从0开始，若要从1开始需要修改
          if(g[u][v]&&!used[v])
          {
              used[v]=true;
              if(linker[v]==-1||dfs(linker[v]))
              {//找增广路，反向
                  linker[v]=u;
                  return true;
              }
          }
        return false;//这个不要忘了，经常忘记这句
    }
    int hungary()
    {
       // puts("hungary");
        int res=0;
        int u;
        memset(linker,-1,sizeof(linker));
        for(u=0;u<uN;u++)
        {
            memset(used,0,sizeof(used));
            if(dfs(u)) res++;
        }
        return res;
    }
int main()
{
    //freopen("cin.txt","r",stdin);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        uN=vN=n;
        memset(g,0,sizeof(g));
        while(m--)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            a--;b--;
            g[a][b]=1;
          //  g[b][a]=1;
        }
        printf("%d\n",n-hungary());
    }
    return 0;
}