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
　

 #include<iostream>

 #include<algorithm>

 #include<vector>

 #include<queue>

 using namespace std;

 const int INF = 0x3f3f3f3f;

 int dis[][];

 int n, m;

 void floyed()

 {

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

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

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

                 if ((dis[i][k] ==  && dis[k][j] == ) || (dis[i][k] ==  && dis[k][j] == ))

                     dis[i][j] = dis[i][k]; //判断ij之间的关系

             }

 }

 int main()

 {

     ios::sync_with_stdio(false);

     while (cin >> n >> m) {

         memset(dis, , sizeof(dis));

         for (int a, b, i = ; i <= m; i++) {

             cin >> a >> b;

             dis[a][b] = ;//a胜b

             dis[b][a] = ;//a输给b

         }

         floyed();

         int ans = ;

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

             int cnt = ;

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

                 if (i == j)continue;

                 if (dis[i][j])cnt++;

             }

             if (cnt == (n - ))ans++;

         }

         cout << ans << endl;

     }

     return ;

 }