#include <bits/stdc++.h>
using namespace std;

const int N = 110, M = 10010;

int n;
int dfn[N], low[N], ts;
int stk[N], top;
int in_stk[N];
int id[N], scc_cnt;
int din[N], dout[N];

// 链式前向星
int e[M], h[N], idx, w[M], ne[M];
void add(int a, int b, int c = 0) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}

// Tarjan算法求强连通分量
void tarjan(int u) {
    dfn[u] = low[u] = ++ts;
    stk[++top] = u;
    in_stk[u] = 1;
    for (int i = h[u]; ~i; i = ne[i]) {
        int v = e[i];
        if (!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if (in_stk[v])
            low[u] = min(low[u], dfn[v]);
    }
    if (dfn[u] == low[u]) {
        ++scc_cnt;
        int x;
        do {
            x = stk[top--];
            in_stk[x] = 0;
            id[x] = scc_cnt;
        } while (x != u);
    }
}

int main() {
    scanf("%d", &n);
    memset(h, -1, sizeof h);

    for (int i = 1; i <= n; i++) {
        int c;
        while (scanf("%d", &c), c) add(i, c); // i支援t
    }

    for (int i = 1; i <= n; i++)
        if (!dfn[i]) tarjan(i);

    // 统计缩点后的DAG，此DAG中出度为0、入度为0的点的个数
    for (int u = 1; u <= n; u++)
        for (int i = h[u]; ~i; i = ne[i]) {
            int v = e[i];
            int a = id[u], b = id[v];
            if (a != b)
                dout[a]++, din[b]++;
        }

    int a = 0, b = 0;
    for (int i = 1; i <= scc_cnt; i++) {
        if (!din[i]) a++;  // 入度为0的强连通块
        if (!dout[i]) b++; // 出度为0的强连通块
    }

    // 输出入度为0的连通块数量,也就是需要软件的数量
    printf("%d\n", a);

    if (scc_cnt == 1) // 如果只有一个强连通块，不用连边
        puts("0");
    else
        printf("%d\n", max(a, b)); // 输出入度为0与出度为0的连通块最大值
    return 0;
}