// Railroad Renovations
// Solution by Jacob Plachta

#include <algorithm>
#include <functional>
#include <numeric>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <complex>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <unordered_set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <sstream>
using namespace std;

#define LL long long
#define LD long double
#define PR pair<int,int>

#define Fox(i,n) for (i=0; i<n; i++)
#define Fox1(i,n) for (i=1; i<=n; i++)
#define FoxI(i,a,b) for (i=a; i<=b; i++)
#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
#define FoxR1(i,n) for (i=n; i>0; i--)
#define FoxRI(i,a,b) for (i=b; i>=a; i--)
#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
#define Min(a,b) a=min(a,b)
#define Max(a,b) a=max(a,b)
#define Sz(s) int((s).size())
#define All(s) (s).begin(),(s).end()
#define Fill(s,v) memset(s,v,sizeof(s))
#define pb push_back
#define mp make_pair
#define x first
#define y second

template<typename T> T Abs(T x) { return(x < 0 ? -x : x); }
template<typename T> T Sqr(T x) { return(x * x); }
string plural(string s) { return(Sz(s) && s[Sz(s) - 1] == 'x' ? s + "en" : s + "s"); }

const int INF = (int)1e9;
const LD EPS = 1e-12;
const LD PI = acos(-1.0);

#define GETCHAR getchar_unlocked

bool Read(int& x)
{
  char c, r = 0, n = 0;
  x = 0;
  for (;;)
  {
    c = GETCHAR();
    if ((c < 0) && (!r))
      return(0);
    if ((c == '-') && (!r))
      n = 1;
    else
      if ((c >= '0') && (c <= '9'))
        x = x * 10 + c - '0', r = 1;
      else
        if (r)
          break;
  }
  if (n)
    x = -x;
  return(1);
}

#define LIM 501

int N, K;
pair<int, PR> O[LIM];

// dyn[i][j] = min. renovations after first i observations (by position),
//             with j rearrangements
int dyn[LIM][LIM];

int ProcessCase()
{
  int i, j, k;
  // input
  Read(N), Read(K);
  Fox(i, N)
  {
    Read(O[i].x), Read(O[i].y.y);
    O[i].y.x = i;
  }
  // sort observations by position
  sort(O, O + N);
  // DP
  Fill(dyn, 60);
  dyn[0][0] = 0;
  Fox(i, N)
  {
    // can start interval here?
    if (i - 1 >= 0 && O[i].x == O[i - 1].x)
      continue;
    // consider all intervals starting here
    int C[2] = { 0 };
    set <PR> S;
    FoxI(j, i, N - 1)
    {
      // update interval info
      S.insert(O[j].y);
      C[O[j].y.y]++;
      // can end interval here?
      if (j + 1 < N && O[j].x == O[j + 1].x)
        continue;
      // compute min. cost for this interval (all 0s before all 1s)
      int cst = min(C[0], C[1]);
      int pc[2] = { 0 };
      for (auto o : S) // consider all 0-1 splitting points
      {
        pc[o.y]++;
        Min(cst, pc[1] + C[0] - pc[0]);
      }
      // update DP values
      FoxI(k, cst, K)
        Min(dyn[j + 1][k], dyn[i][k - cst] + (C[1] > 0));
    }
  }
  // compute answer
  int ans = INF;
  Fox(k, K + 1)
    Min(ans, dyn[N][k]);
  return(ans == INF ? -1 : ans);
}

int main()
{
  int T, t;
  Read(T);
  Fox1(t, T)
    printf("Case #%d: %d\n", t, ProcessCase());
  return(0);
}