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// Ethan Traverses a Tree
// Solution by Jacob Plachta

#define DEBUG 0

#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 <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);

#if DEBUG
#define GETCHAR getchar
#else
#define GETCHAR getchar_unlocked
#endif

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 2000

int A[LIM],B[LIM];
int cnt[2],ord[2][LIM];

void rec(int z,int i)
{
		if (i<0)
			return;
		if (!z)
			ord[z][cnt[z]++]=i;
	rec(z,A[i]);
	rec(z,B[i]);
		if (z)
			ord[z][cnt[z]++]=i;
}

int main()
{
		if (DEBUG)
			freopen("in.txt","r",stdin);
	// vars
	int T,t;
	int N,K;
	int i,j,k,a,b;
	vector<int> con[LIM];
	int L[LIM];
	queue<int> Q;
	// testcase loop
	Read(T);
		Fox1(t,T)
		{
			// input
			Read(N),Read(K);
				Fox(i,N)
					Read(A[i]),Read(B[i]),A[i]--,B[i]--;
			// compute both traversals
			Fill(cnt,0);
				Fox(i,2)
					rec(i,0);
			// construct graph of nodes with equal labels
				Fox(i,N)
					con[i].clear();
				Fox(i,2)
					Fox(j,N)
						con[ord[i][j]].pb(ord[1-i][j]);
			// split graph into components
			k=0;
			Fill(L,-1);
				Fox(i,N)
					if (L[i]<0)
					{
							if (k<K)
								k++;
						Q.push(i),L[i]=k;
							while (!Q.empty())
							{
								j=Q.front(),Q.pop();
									Fox(a,Sz(con[j]))
										if (L[b=con[j][a]]<0)
											Q.push(b),L[b]=k;
							}
					}
			// output
			printf("Case #%d:",t);
				if (k<K)
					printf(" Impossible\n");
				else
				{
						Fox(i,N)
							printf(" %d",L[i]);
					printf("\n");
				}
		}
	return(0);
}