#include <bits/stdc++.h>
using namespace std;
static const int MAXN = 500;
int N, K;
vector<int> adj[MAXN];
int dp2[MAXN][MAXN][2]; // classical DP: max edges from st, using only vertices in open arc (st->en,dir)
int dp1[MAXN][MAXN][2]; // one-crossing DP: max edges from st with at most one crossing of chord en->st
inline bool inrange(int st, int en, int dir, int x) {
if (x == st || x == en) return false;
if (dir == 0) {
// CCW from st to en
int d_se = (en - st + N) % N;
int d_sx = (x - st + N) % N;
return (d_sx > 0 && d_sx < d_se);
} else {
// CW from st to en
int d_se = (st - en + N) % N;
int d_sx = (st - x + N) % N;
return (d_sx > 0 && d_sx < d_se);
}
}
// DP for k=0: longest path without crossing chord en->st
int dfs2(int st, int en, int dir) {
int &res = dp2[st][en][dir];
if (res != -1) return res;
int best = 0;
for (int nxt : adj[st]) {
if (nxt == en || inrange(st, en, dir, nxt)) {
best = max(best, 1 + dfs2(nxt, en, dir));
}
}
return res = best;
}
// DP for k=1: longest path with at most one crossing of chord en->st
int dfs1(int st, int en, int dir) {
int &res = dp1[st][en][dir];
if (res != -1) return res;
// base case: no moves when start equals end
if (st == en) return res = 0;
int &res = dp1[st][en][dir];
if (res != -1) return res;
// Option 1: no crossing at all
int best = dfs2(st, en, dir);
// Option 2: take one step
for (int nxt : adj[st]) {
if (nxt == en || inrange(st, en, dir, nxt)) {
// still inside interior arc, crossing not used
best = max(best, 1 + dfs1(nxt, en, dir));
} else {
// this edge crosses chord en->st; after this, no more crossings
best = max(best, 1 + dfs2(nxt, en, 1 - dir));
}
}
return res = best;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> N >> K;
for (int i = 0; i < N; i++) {
adj[i].clear();
while (true) {
int x; cin >> x;
if (x == 0) break;
adj[i].push_back(x - 1);
}
}
memset(dp2, -1, sizeof(dp2));
memset(dp1, -1, sizeof(dp1));
int answer = 0;
int startHarbor = 1;
for (int s = 0; s < N; s++) {
for (int t : adj[s]) {
for (int dir = 0; dir < 2; dir++) {
int val;
if (K == 0) {
val = 1 + dfs2(t, s, 1 - dir);
} else {
val = 1 + dfs1(t, s, 1 - dir);
}
if (val > answer) {
answer = val;
startHarbor = s + 1;
}
}
}
}
cout << answer << "\n" << startHarbor << "\n";
return 0;
}
