/*
-We need 3 dp tables to figure out the answer:
-dp1[l][r][a] - Max length of a path starting at l and ending at r, going in direction a.
-dp2[l][r][a] - Max length of a path starting at l, visiting r and then continuing from r in the opposite direction(still in interval [l,r] and it can swap the direction multiple times)
-dp3[l][r][a] - Max of dp2[l][l...r][a] : We need this to calculate the case where we have an intersection in O(n^3) and not O(n^4)
-The case where there is no crossing can be solved using only the first 2 dps, the second case requires us to go though all the states of dp1, and for every one of them(if there is a path from l to r)
we need to try to elongate it such that we only cross once at the starting segment(we first add a starting segment and then add an edge crossing it, followed by a path of maximal length from that edge to harbors not in the interval.
*/
#include <bits/stdc++.h>
using namespace std;
const int N=501;
bool adj[N][N];
int dp1[N][N][2],dp2[N][N][2],dp3[N][N][2],b[N][2];
pair<int,int> ans;
void calc(int l,int r,int a)
{
if(adj[l][r])
{
dp1[l][r][a]=1;
dp2[l][r][a]=1+dp3[r][b[l][a]][a^1];
}
else
{
dp1[l][r][a]=dp2[l][r][a]=INT_MIN/2;
}
for(int k=b[l][a];k!=r;k=b[k][a])
{
dp1[l][r][a]=max(dp1[l][r][a],dp1[l][k][a]+dp1[k][r][a]);
dp2[l][r][a]=max(dp2[l][r][a],dp1[l][k][a]+dp2[k][r][a]);
}
dp3[l][r][a]=max(dp3[l][b[r][a^1]][a],dp2[l][r][a]);
}
int main()
{
int n,k;
scanf("%i %i",&n,&k);
for(int i=0;i<n;i++)
{
int t;
scanf("%i",&t);
while(t!=0)
{
adj[i][t-1]=true;
scanf("%i",&t);
}
b[i][0]=(i+1)%n;
b[i][1]=(i-1+n)%n;
}
for(int l=1;l<n;l++)
for(int i=0;i<n;i++)
calc(i,(i+l)%n,0),calc((i+l)%n,i,1);
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
for(int a=0;a<2;a++)
ans=max(ans,make_pair(dp2[i][j][a],i));
if(k)
{
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
for(int a=0;a<2;a++)
if(dp1[i][j][a]>0)
{
int k=b[j][a];
for(;k!=i;k=b[k][a])
if(adj[k][i])
break;
if(k!=i)
for(int l=b[k][a];l!=i;l=b[l][a])
if(adj[j][l])
ans=max(ans,make_pair(max(dp3[l][b[i][a^1]][a],dp3[l][b[k][a]][a^1])+dp1[i][j][a]+2,k));
}
}
printf("%i\n%i\n",ans.first,ans.second+1);
return 0;
}
// :')
