#include <stdio.h>

 #include <stdlib.h>

 #include <stdbool.h>

 #define maxn 1000

 //if a beats b and b beats c , then a beats c

 int main()

 {

     long n,m,i,j,k,a,b,ans,total=;

     bool vis[maxn][maxn];

     scanf("%ld%ld",&n,&m);

     for (i=;i<=n;i++)

         for (j=;j<=n;j++)

             vis[i][j]=false;

     for (i=;i<=m;i++)

     {

         scanf("%ld%ld",&a,&b);

         vis[a][b]=true;

     }

     for (i=;i<=n;i++)

         for (j=;j<=n;j++)

                 for (k=;k<=n;k++)

                     vis[j][k]=vis[j][k] || (vis[j][i] && vis[i][k]);

     //to cow i, if all the result of content(n-1) has been known,

     //then the rank can be determined

     for (i=;i<=n;i++)

     {

         ans=;

         for (j=;j<=n;j++)

             if (vis[i][j] || vis[j][i])

                 ans++;

         if (ans==n-)

             total++;

     }

     printf("%ld\n",total);

     return ;

 }

 #include <stdio.h>

 #include <stdlib.h>

 #include <stdbool.h>

 #define maxn 1000

 //n个点，点编号为1~n，m条边

 //传递闭包:判断图中任意两点是否可达

 //时间复杂度:n*n*n

 int main()

 {

     long n,m,i,j,k,a,b;

     bool vis[maxn][maxn];

     scanf("%ld%ld",&n,&m);

     for (i=;i<=n;i++)

         for (j=;j<=n;j++)

             vis[i][j]=false;

     for (i=;i<=m;i++)

     {

         scanf("%ld%ld",&a,&b);

         vis[a][b]=true;

     }

     //如果是无向图，a能到达b意味着b也能到达a，所以可以求a->b(a<b),i<j<k

     //如果没有if:操作n*n*n

     //如果有if:判断n*n+n*(n-1)*n次，操作n*(n-1)*(n-2)

     //所以直接不用if

     //从j点到k点：从j到k经过的点的编号小于等于i(i=1,2,…,n)。当然也可以从j直达到k。

     for (i=;i<=n;i++)

         for (j=;j<=n;j++)

             //if (i!=j)

                 for (k=;k<=n;k++)

                     //if (i!=j && i!=k)

                     vis[j][k]=vis[j][k] || (vis[j][i] && vis[i][k]);

                         //if (vis[j][i] && vis[i][k])

                             //vis[j][k]=true;

     for (i=;i<=n;i++)

     {

         printf("%ld : ",i);

         for (j=;j<=n;j++)

             if (vis[i][j])

                 printf("%ld ",j);

         printf("\n");

     }

     return ;

 }

 /*

 5 5

 1 2

 2 3

 1 3

 4 5

 5 3

 */