N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

* Line 1: A single integer representing the number of cows whose ranks can be determined
　

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 105

#define eps 1e-9

const int inf=0x7fffffff;   //无限大

int g[maxn][maxn];

int main()

{

    int n,m;

    while(cin>>n>>m)

    {

        int a,b;

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

        {

            cin>>a>>b;

            g[a][b]=;

        }

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

        {

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

            {

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

                {

                    if(g[i][k]&&g[k][j])

                        g[i][j]=;

                }

            }

        }

        int ans=;

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

        {

            int flag=;

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

            {

                if(i==j)

                    continue;

                if(g[i][j]==&&g[j][i]==)

                {

                    flag=;

                    break;

                }

            }

            if(flag)

                ans++;

        }

        cout<<ans<<endl;

    }

}