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// Ethan Finds the Maximum Subarray Sum
// 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);
}

int main()
{
		if (DEBUG)
			freopen("in.txt","r",stdin);
	// vars
	int T,t;
	int N,K;
	int i,m,w,s,c,ans;
	int A[100];
	// testcase loop
	Read(T);
		Fox1(t,T)
		{
			// input
			Read(N),Read(K);
			N=N*2-1;
			m=0;
				Fox(i,N)
					if (i%2==0)
						Read(A[i]),Max(m,A[i]);
				if (A[N-1]<0)
					N--;
			// consider each possible value w for Ethan's answer
			ans=0;
				FoxI(w,m,N*K+1)
				{
					// greedily fill in values
					s=c=0;
						FoxI(i,(A[0]<0 ? 1 : 0),N-1)
							if (i%2==0) // forced value?
							{
								s+=A[i];
									if (A[i]>=0)
										c+=A[i];
									else
										c=0;
							}
							else
							if (i==N-1) // last value?
							{
								if (c+K<=w) // fill with K if possible
									s+=K,c+=K;
							}
							else
							if (((A[i+1]<0) && (c+K<=w)) || ((A[i+1]>=0) && (c+K+A[i+1]<=w))) // can fill with K?
								s+=K,c+=K;
							else
								s--,c=0; // fill with -1 otherwise
					Max(ans,s-w);
				}
			// output
			printf("Case #%d: %d\n",t,ans);
		}
	return(0);
}