#include <bits/stdc++.h>
using namespace std;
static const int MAXN = 500;
int N, K;
vector<int> adj[MAXN];
int f2_dp[MAXN][MAXN][2]; // classical DP: max edges from st to en, no crossing
char vis2[MAXN][MAXN][2];
int f1_dp[MAXN][MAXN][2]; // one-crossing DP: max edges from st to en, at most one crossing
char vis1[MAXN][MAXN][2];
// Check if x lies strictly between st and en along the circle in direction dir
inline bool inrange(int st, int en, int dir, int x) {
if (x == st || x == en) return false;
if (dir == 0) {
int d_se = (en - st + N) % N;
int d_sx = (x - st + N) % N;
return (d_sx > 0 && d_sx < d_se);
} else {
int d_se = (st - en + N) % N;
int d_sx = (st - x + N) % N;
return (d_sx > 0 && d_sx < d_se);
}
}
// Classical DP: no crossings allowed
int dfs2(int st, int en, int dir) {
int &res = f2_dp[st][en][dir];
if (vis2[st][en][dir]) return res;
vis2[st][en][dir] = 1;
if (st == en) {
return res = 0;
}
int best = -1000000000;
for (int nxt : adj[st]) {
if (nxt == en || inrange(st, en, dir, nxt)) {
int tmp = dfs2(nxt, en, dir);
if (tmp >= 0) best = max(best, tmp + 1);
}
}
return res = best;
}
// One-crossing DP: at most one crossing with chord st-en
int dfs1(int st, int en, int dir) {
int &res = f1_dp[st][en][dir];
if (vis1[st][en][dir]) return res;
vis1[st][en][dir] = 1;
// Case 0: no crossing at all
int best = dfs2(st, en, dir);
// Try transitions
for (int nxt : adj[st]) {
if (nxt == en || inrange(st, en, dir, nxt)) {
// no crossing on this edge
int tmp = dfs1(nxt, en, dir);
if (tmp >= 0) best = max(best, tmp + 1);
} else {
// crossing occurs on this edge
int tmp = dfs2(nxt, en, 1 - dir);
if (tmp >= 0) best = max(best, tmp + 1);
}
}
return res = best;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> N >> K;
for (int i = 0; i < N; i++) {
while (true) {
int x; cin >> x;
if (x == 0) break;
adj[i].push_back(x - 1);
}
}
// Initialize DP visited arrays
memset(vis2, 0, sizeof(vis2));
memset(vis1, 0, sizeof(vis1));
int answer = 0;
int startHarbor = 1;
// Precompute all f2 and f1 states on demand
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 = dfs2(t, s, dir);
} else {
val = dfs1(t, s, dir);
}
if (val >= 0) val += 1; // add the first stage s->t
if (val > answer) {
answer = val;
startHarbor = s + 1;
}
}
}
}
// Output
cout << answer << '\n' << startHarbor << '\n';
return 0;
}
