1 //2017-08-20
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <iostream>
  5 #include <algorithm>
  6 #include <vector>
  7 
  8 using namespace std;
  9 
 10 const int N = 10010;
 11 vector<int> G[N];//邻接表存图
 12 vector<int> rG[N];//存反向图
 13 vector<int> vs;//后序遍历顺序的顶点列表
 14 bool vis[N];
 15 int cmp[N];//所属强连通分量的拓扑序
 16 
 17 void add_edge(int u, int v){
 18     G[u].push_back(v);
 19     rG[v].push_back(u);
 20 }
 21 
 22 //input: u 顶点
 23 //output: vs 后序遍历顺序的顶点列表
 24 void dfs(int u){
 25     vis[u] = true;
 26     for(int i = 0; i < G[u].size(); i++){
 27         int v = G[u][i];
 28         if(!vis[v])
 29               dfs(v);
 30     }
 31     vs.push_back(u);
 32 }
 33 
 34 //input: u 顶点编号; k 拓扑序号
 35 //output: cmp[] 强连通分量拓扑序
 36 void rdfs(int u, int k){
 37     vis[u] = true;
 38     cmp[u] = k;
 39     for(int i = 0; i < rG[u].size(); i++){
 40         int v = rG[u][i];
 41         if(!vis[v])
 42               rdfs(v, k);
 43     }
 44 }
 45 
 46 //Strongly Connected Component 强连通分量
 47 //input: n 顶点个数
 48 //output: k 强连通分量数;
 49 int scc(int n){
 50     memset(vis, 0, sizeof(vis));
 51     vs.clear();
 52     for(int u = 0; u < n; u++)
 53         if(!vis[u])
 54               dfs(u);
 55     int k = 0;
 56     memset(vis, 0, sizeof(vis));
 57     for(int i = vs.size()-1; i >= 0; i--)
 58           if(!vis[vs[i]])
 59               rdfs(vs[i], k++);
 60     return k;
 61 }
 62 
 63 void solve(int n){
 64     int k = scc(n);
 65     int u = 0, ans = 0;
 66     for(int v = 0; v < n; v++){
 67         if(cmp[v] == k-1){
 68             u = v;
 69             ans++;
 70         }
 71     }
 72     memset(vis, 0, sizeof(vis));
 73     rdfs(u, 0);
 74     for(int i = 0; i < n; i++){
 75         if(!vis[i]){
 76             ans = 0;
 77             break;
 78         }
 79     }
 80     printf("%d\n", ans);
 81 }
 82 
 83 int main()
 84 {
 85     int n, m;
 86     while(scanf("%d%d", &n, &m)!=EOF){
 87         int u, v;
 88         for(int i = 0; i < n; i++){
 89             G[i].clear();
 90             rG[i].clear();
 91         }
 92         while(m--){
 93             scanf("%d%d", &u, &v);
 94             u--; v--;
 95             add_edge(u, v);
 96         }
 97         solve(n);
 98     }
 99 
100     return 0;
101 }