C

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.

Input

* 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

Output

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

Sample Input

5 54 34 23 21 22 5

Sample Output

2   

给你n只牛，每只牛的技能的强弱不一样。现在给你m组牛的胜负关系，问你可以确定多少只牛的技能的排名。

先用floyd将存在的间接关系变为直接关系，例如i和k有关系，k和j有关系，那么通过floyd可以直接得到i和j也有关系。然后再遍历每一只牛，如果他和每一只牛的关系都确定了，那么他的排名必然就能确定        

#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<vector>using namespace std;int g[110][110],out[110];int n,m;void floyd(){    for(int k=1;k<=n;k++)    {        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)                g[i][j]=g[i][j]||(g[i][k]&&g[k][j]);        }    }}int main(){    ///freopen("1.txt","r",stdin);    while(cin>>n>>m)    {        memset(g,0,sizeof(g));        for(int i=1;i<=m;i++)        {            int x,y;cin>>x>>y;            g[x][y]=1;        }        floyd();        int co=0;        for(int i=1;i<=n;i++)        {            int flag=0;            for(int j=1;j<=n;j++)                if(g[i][j]||g[j][i])flag++;            if(flag==n-1)co++;        }        cout<<co<<endl;    }}