// 题目给的 M 为 1e6 但是我们要建两个图 所以 设 M 为 2e6
const int N = 1e5 + 10, M = 2e6 + 10;
int n, m, mod;
int h[N],hs[N],e[M],ne[M],idx;
int low[N],dfn[N];
int scc,timestamp;
stack<int>stk;
bool in_stk[N];
int id[N],siz[N];
int f[N],g[N];
unordered_set<LL>S;
void add(int h[],int a,int b){
  e[idx] = b,ne[idx] = h[a],h[a] = idx++;
}
void tarjan(int u){
  dfn[u] = low[u] = ++timestamp;
  stk.push(u),in_stk[u] = true;
  for(int i = h[u];~i;i = ne[i]){
    int j = e[i];
    if(!dfn[j]){
      tarjan(j);
      low[u] = min(low[u],low[j]);
    }
    else if(in_stk[j]){
      low[u] = min(low[u],dfn[j]);
    }
  }
  if(dfn[u] == low[u]){
    int y = 0;
    ++scc;
    do{
      y = stk.top();
      stk.pop();
      in_stk[y] = false;
      siz[scc]++;
      id[y] = scc;
    }while(y != u);
  }
}
void solve() {
  scanf("%d%d%d",&n,&m,&mod);
  memset(h,-1,sizeof h);
  memset(hs,-1,sizeof hs);
  while(m--){
    int a,b;scanf("%d%d",&a,&b);
    add(h,a,b);
  }
  for(int i = 1;i <= n;++i){
    if(!dfn[i])
      tarjan(i);
  }
  for(int i = 1; i <= n;++i){ // 对scc建图
    for(int j = h[i];~j;j = ne[j]){
      int k = e[j];
      int a = id[i], b = id[k];
      LL hash = a * 1000000ll + b;
      if(a != b && !S.count(hash)){
        add(hs,a,b);
        S.insert(hash);
      }
    }
  }
  for(int i = scc;i;--i){ // scc递减的顺序为 拓扑序
    if(!f[i]){
      f[i] = siz[i];
      g[i] = 1;
    }
    for(int j = hs[i];~j;j = ne[j]){
      int k = e[j];
      if(f[k] < f[i] + siz[k]){
        f[k] = f[i] + siz[k];
        g[k] = g[i];
      }
      else if(f[k] == f[i] + siz[k]){
        g[k] = (g[k] + g[i]) % mod;
      }
    }
  }
  LL sum = 0,maxn = 0;
  for(int i = 1; i <= scc;++i){
    if(f[i] > maxn){
      maxn = f[i];
      sum = g[i];
    }
    else if(f[i] == maxn){
      sum = (sum + g[i]) % mod;
    }
  }
  printf("%lld\n",maxn);
  printf("%lld\n",sum);
}
signed main() {
    //int _; cin >> _;
    //while (_--)
        solve();
    return 0;
}