/*
    求最小加邊使有向圖變為強連通
    先縮點，然后計算各個強連通分量的入度為0的個數，出度為0的個數
    1) SCC=1, 不需添邊
    2) 添max( incnt, outcnt )   
                    貪心策略，匯基向點基連邊，再添邊的連接剩下的點基/匯基即可

*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;

const int MAXN=20010;

int n,m;
vector<int>G[2][MAXN];
int ID[MAXN],SCC,flag;
int order[MAXN],cnt;
int in[MAXN],out[MAXN];

void dfs(int u){
    ID[u]=SCC;
    for(int i=0;i<G[flag][u].size();i++){
        int v=G[flag][u][i];
        if(!ID[v])dfs(v);
    }
    if(flag)order[++cnt]=u;
}

int main(){
    int T,a,b;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++){
            G[0][i].clear();
            G[1][i].clear();
        }
        while(m--){
            scanf("%d%d",&a,&b);
            G[0][a].push_back(b);
            G[1][b].push_back(a);
        }
        flag=SCC=1;
        cnt=0;
        memset(ID,0,sizeof(ID));
        for(int i=1;i<=n;i++)
            if(!ID[i])dfs(i);
        flag=SCC=0;
        memset(ID,0,sizeof(ID));
        for(int i=cnt;i;i--){//注意是order[i]
            if(ID[order[i]])continue;
            SCC++;
            dfs(order[i]);
        }
        if(SCC==1){puts("0");continue;}
        memset(out,0,sizeof(out));
        memset(in,0,sizeof(in));
        for(int i=1;i<=n;i++)
            for(int j=0;j<G[0][i].size();j++){
                int v=G[0][i][j];
                if(ID[i]!=ID[v]){
                    in[ID[v]]++;
                    out[ID[i]]++;
                }
            }
        int incnt=0,outcnt=0;
        for(int i=1;i<=SCC;i++){
            if(in[i]==0)incnt++;
            if(out[i]==0)outcnt++;
        }
        printf("%d\n",max(incnt,outcnt));
    }
    return 0;
}