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Description

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 5
4 3
4 2
3 2
1 2
2 5

Sample Output

2

// 牛 A能打败 牛B 问有多少个牛的顺序能被确定。

//记录每个牛的度，如果等于N-1 表示能确定顺序。

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#include 
  
   
#include 
   
     #include 
    
      #include 
     
       #include 
      
        #include 
       
         #define N 109 using namespace std; int mp[N][N]; int n,m; int a,b; int deg[N]; int main() { while(~scanf("%d %d",&n,&m)) { memset(deg,0,sizeof deg); memset(mp,0,sizeof mp); for(int i=1;i<=m;i++) { scanf("%d %d",&a,&b); mp[a][b]=1; } for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) for(int k=1;k<=n;k++) if(mp[j][i] && mp[i][k]) mp[j][k]=1; for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) { if(mp[i][j]) { deg[i]++; deg[j]++; } } int ans=0; for(int i=1;i<=n;i++) if(deg[i]==n-1) ans++; printf("%d\n",ans); } return 0; } 
       
      
     
    
   
  

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