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 <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
using namespace std;int N, M;
int mp[110][110];void floyd() {for(int k = 1; k <= N; k ++) {for(int i = 1; i <= N; i ++) {if(mp[i][k])for(int j = 1; j <= N; j ++) {if(mp[k][j] == 1 && mp[i][k] == 1) {mp[i][j] = 1;mp[j][i] = -1;}else if(mp[k][j] == -1 && mp[i][k] == -1) {mp[i][j] = -1;mp[j][i] = 1;}else continue;}}}
}int main() {memset(mp, 0, sizeof(mp));scanf("%d%d", &N, &M);for(int i = 1; i <= M; i ++) {int a, b;scanf("%d%d", &a, &b);mp[a][b] = 1;mp[b][a] = -1;}floyd();int ans = 0;for(int i = 1; i <= N; i ++) {int sum = 0;for(int j = 1; j <= N; j ++)if(mp[i][j])sum ++;if(sum == N - 1)ans ++;}printf("%d\n", ans);return 0;
}