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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110, M = 10010;
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int n;
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int dfn[N], low[N], ts;
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int stk[N], top;
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int in_stk[N];
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int id[N], scc_cnt;
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int din[N], dout[N];
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// 链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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// Tarjan算法求强连通分量
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void tarjan(int u) {
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dfn[u] = low[u] = ++ts;
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stk[++top] = u;
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in_stk[u] = 1;
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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if (!dfn[v]) {
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tarjan(v);
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low[u] = min(low[u], low[v]);
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} else if (in_stk[v])
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low[u] = min(low[u], dfn[v]);
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}
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if (dfn[u] == low[u]) {
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++scc_cnt;
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int x;
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do {
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x = stk[top--];
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in_stk[x] = 0;
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id[x] = scc_cnt;
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} while (x != u);
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}
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}
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int main() {
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scanf("%d", &n);
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memset(h, -1, sizeof h);
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for (int i = 1; i <= n; i++) {
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int c;
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while (scanf("%d", &c), c) add(i, c); // i支援t
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}
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for (int i = 1; i <= n; i++)
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if (!dfn[i]) tarjan(i);
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// 统计缩点后的DAG，此DAG中出度为0、入度为0的点的个数
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for (int u = 1; u <= n; u++)
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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int a = id[u], b = id[v];
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if (a != b)
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dout[a]++, din[b]++;
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}
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int a = 0, b = 0;
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for (int i = 1; i <= scc_cnt; i++) {
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if (!din[i]) a++; // 入度为0的强连通块
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if (!dout[i]) b++; // 出度为0的强连通块
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}
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// 输出入度为0的连通块数量,也就是需要软件的数量
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printf("%d\n", a);
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if (scc_cnt == 1) // 如果只有一个强连通块，不用连边
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puts("0");
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else
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printf("%d\n", max(a, b)); // 输出入度为0与出度为0的连通块最大值
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return 0;
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}
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