#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 dp[2][2][501][501];
int N,k;
int nor(int x) { x = (x%N+N)%N; if (x == 0) x += N; return x; }
void MX(int& a, int b) { a = max(a,b); }
pi solve0() {
FOR(len,2,N) FOR(i,1,N+1) {
int en = nor(i+len);
FOR(j,1,len) {
int ex = max(dp[0][0][nor(i+j)][en],dp[0][1][i][nor(i+j)])+1;
if (ok[i][nor(i+j)]) MX(dp[0][0][i][en],ex);
if (ok[en][nor(i+j)]) MX(dp[0][1][i][en],ex);
}
}
pi ret = {0,0};
FOR(i,1,N+1) FOR(j,1,N+1) if (ok[i][j])
ret = max(ret,{max(dp[0][1][i][j],dp[0][0][j][i])+1,i});
return ret;
}
pi solve1() {
FOR(len,1,N) FOR(i,1,N+1) {
int en = nor(i+len);
dp[1][1][i][en] = dp[1][0][i][en] = -MOD;
F0R(j,len+1) {
if (ok[nor(i+j)][en]) MX(dp[1][1][i][en],dp[1][1][i][nor(i+j)]+1);
if (ok[nor(i+j)][i]) MX(dp[1][0][i][en],dp[1][0][nor(i+j)][en]+1);
}
}
pi ret = {0,0};
FOR(a,1,N+1) FOR(len,1,N+1) if (ok[a][nor(a+len)]) {
vi l, r;
FOR(b,1,N+1) if (b != len) {
if (b < len) l.pb(nor(a+b));
else r.pb(nor(a+b));
}
for (int i: l) for (int j: r) if (ok[i][j])
ret = max(ret,{1+dp[1][0][i][nor(a+len)]+1+max(dp[0][0][j][a],dp[0][1][nor(a+len)][j]),a});
for (int i: r) for (int j: l) if (ok[i][j])
ret = max(ret,{1+dp[1][1][nor(a+len)][i]+1+max(dp[0][0][j][nor(a+len)],dp[0][1][a][j]),a});
}
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) t = max(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
*/
