#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) {
if (x <= 0) x += N;
if (x > N) x -= N;
return x;
}
template<class T> void MX(T& a, T b) { a = max(a,b); }
int alt[2][501][501];
pi ret = {0,-1};
void 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);
}
}
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});
}
int no[2][501][501];
void test2(int len, int i) {
int en = nor(i+len);
for (int j = 0; j < len; ++j) {
int m = nor(i+j);
if (j && ok[m][i]) {
pi t = {no[0][i][en]+max(alt[1][m][en],alt[0][en][i])+1,-m};
// if (t.f == 194) cout << len << " " << i << " " << no[0][i][en] << " " << alt[1][m][en] << " " << alt[0][en][i] << " " << t.f << " " << t.s << "\n";
MX(ret,t);
}
if (ok[m][en]) MX(no[0][i][en],no[0][i][m]+1);
}
}
void test4(int len, int i) {
int st = nor(i-len);
for (int j = 0; j < len; ++j) {
int m = nor(i-j);
if (j && ok[m][i]) MX(ret,{no[1][st][i]+max(alt[0][st][m],alt[1][i][st])+1,-m});
if (ok[m][st]) MX(no[1][st][i],no[1][m][i]+1);
}
}
void solve1() {
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;
}
FOR(len,1,N) FOR(i,1,N+1) {
test2(len,i);
test4(len,i);
}
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> N >> k;
// FOR(i,1,N+1) FOR(j,1,N+1) if (i != j) ok[i][j] = rand() % 2;
FOR(i,1,N+1) {
int x;
while (cin >> x) {
if (x == 0) break;
if (i != x) ok[i][x] = 1;
}
}
solve0();
//cout << "HA " << alt[1][133][126] << "\n";
if (k == 1) solve1();
cout << ret.f << "\n" << -ret.s << "\n";
//FOR(i,1,N+1) if (ok[127][i] && ok[i][126]) cout << i << "\n";
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
