#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
bool ok[501][501];
int N,k;
int nor(int x) { x = (x%N+N)%N; if (x == 0) x += N; return x; }
template<class T> void MX(T& a, T b) { a = max(a,b); }
int alt[2][501][501];
pi solve0() {
FOR(len,2,N) FOR(i,1,N+1) {
int en = nor(i+len);
FOR(j,1,len) {
int m = nor(i+j);
int ex = max(alt[0][m][en],alt[1][i][m])+1;
if (ok[i][m]) MX(alt[0][i][en],ex);
if (ok[en][m]) MX(alt[1][i][en],ex);
}
}
pi ret = {0,0};
FOR(i,1,N+1) FOR(j,1,N+1) if (ok[i][j])
MX(ret,{max(alt[1][i][j],alt[0][j][i])+1,i});
return ret;
}
int no[2][501][501];
pi yes[2][501][501];
pi solve1() {
pi ret = {0,0};
F0R(i,2) FOR(j,1,N+1) FOR(k,1,N+1) {
if (j != k) no[i][j][k] = -MOD;
else no[i][j][k] = 0;
yes[i][j][k] = {-MOD,-MOD};
}
FOR(len,1,N) FOR(i,1,N+1) {
int en = nor(i+len);
int existsEdge = 0;
for (int j = len-1; j >= 0; --j) {
int m = nor(i+j);
if (ok[m][en]) {
MX(no[0][i][en],no[0][i][m]+1);
if (existsEdge) MX(yes[0][i][en],{no[0][i][en]+1,existsEdge});
}
if (ok[m][i]) existsEdge = m;
}
MX(ret,{yes[0][i][en].f+alt[0][en][i]+1,yes[0][i][en].s});
no[0][i][en] = -MOD;
for (int j = 0; j < len; ++j) {
int m = nor(i+j);
if (j) MX(ret,{no[0][i][en]+alt[1][m][en]+1,m});
if (ok[m][en]) MX(no[0][i][en],no[0][i][m]+1);
}
int st = nor(i-len);
existsEdge = 0;
for (int j = len-1; j >= 0; --j) {
int m = nor(i-j);
if (ok[m][st]) {
MX(no[1][st][i],no[1][m][i]+1);
if (existsEdge) MX(yes[1][st][i],{no[1][st][i]+1,existsEdge});
}
if (ok[m][i]) existsEdge = m;
}
MX(ret,{yes[1][st][i].f+alt[1][i][st]+1,yes[1][st][i].s});
no[1][st][i] = -MOD;
for (int j = 0; j < len; ++j) {
int m = nor(i-j);
if (j) MX(ret,{no[1][st][i]+alt[0][st][m]+1,m});
if (ok[m][st]) MX(no[1][st][i],no[1][m][i]+1);
}
}
return ret;
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> N >> k;
FOR(i,1,N+1) {
int x;
while (cin >> x) {
if (x == 0) break;
ok[i][x] = 1;
}
}
pi t = solve0();
if (k == 1) MX(t,solve1());
cout << t.f << "\n" << t.s;
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
