/**

 * CTU Open 2017

 * Problem Solution: Ice cream samples

 * Idea: Use two pointers to consecutively enlarge/shorten result interval - linear time solution

 * @author Josef Malik

 */

 

#include <algorithm>

#include <cmath>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <iostream>

#include <sstream>

#include <map>

#include <set>

#include <queue>

#include <vector>

 

using namespace std;

 

typedef long long int ll;

typedef pair<int, int> pii;

 

#define PB push_back

#define MP make_pair

 

#define FOR(prom, a, b) for(int prom = (a); prom < ((int)(b)); prom++)

#define FORD(prom, a, b) for(int prom = (a); prom > (b); prom--)

#define FORDE(prom, a, b) for(int prom = (a); prom >= (b); prom--)

#define R1(a) do{scanf("%d", &(a));}while(0)

#define R2(a, b) do{scanf("%d%d", &(a), &(b));}while(0)

#define R3(a, b, c) do{scanf("%d%d%d", &(a), &(b), &(c));}while(0)

#define SV(vec) do{int s_v_;scanf("%d", &(s_v_));vec.PB(s_v_);}while(0)

#define MM(co, cim) memset((co), (cim), sizeof((co)))

#define DEB(x) cerr << ">>> " << #x << " : " << x << endl;

#define INF 1000000007

 

int n, k, li, p1, p2, ct[1000014], ctu, ctc, res;

vector<int> st[1000014];

 

int main ()

{

  while (scanf("%d%d", &n, &k) != -1)

  {

    FOR(i, 0, n) st[i].clear();

    FOR(i, 0, n)

    {

      R1(li);

      FOR(j, 0, li) SV(st[i]);

    }

    res = INF;

    MM(ct, 0);

    p1 = ctc = ctu = 0;

    p2 = -1;

    while (p1 < 2 * n)

    {

      if (p2 == 2 * n - 1 || p2 - p1 + 1 >= n || ctu == k)

      {

        if (ctu == k) res = min(res, ctc);

        FOR(i, 0, st[p1 % n].size())

        {

          if (--ct[st[p1 % n][i]]);else --ctu;

          --ctc;

        }

        ++p1;

      }

      else

      {

        ++p2;

        FOR(i, 0, st[p2 % n].size())

        {

          if (++ct[st[p2 % n][i]] == 1) ++ctu;

          ++ctc;

        }

      }

    }

    printf("%d\n", res == INF ? -1 : res);

  }

   

  return 0;

}