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
　

//Asimple
#include <iostream>
#include <algorithm>
#define mod 100000
#define CLS(a, v) memset(a, v, sizeof(a))
#define debug(a)  cout << #a << " = "  << a <<endl
#define dobug(a, b)  cout << #a << " = "  << a << " " << #b << " = " << b << endl
using namespace std;
typedef long long ll;
const int maxn = 500+5;const int INF = (1 << 16);int n, m, num, T, k, len, ans, sum, x, y, z;
int Map[maxn][maxn];
void solve(){for(int k=1; k<=n; k++)for(int i=1; i<=n; i++)for(int j=1; j<=n; j++)if( Map[i][k] && Map[k][j] )Map[i][j] = 1;/*传递闭包  只有这个点和其余所有的点的关系都是确定的这个点才是确定的*/ans = 0;int j;for(int i=1; i<=n; i++) {for(j=1; j<=n; j++) {if( i==j ) continue;if( Map[i][j]==0 && Map[j][i]==0) break;}if( j>n ) ans ++;}cout << ans << endl;
}void input() {ios_base::sync_with_stdio(false);while( cin >> n >> k ) {CLS(Map, 0);while( k -- ) {cin >> x >> y;Map[x][y] = 1;}solve();}
}int main(){input();return 0;
}