Bear has a large, empty ground for him to build a home. He decides to build a row of houses, one after another, say n in total.

The houses are designed with different height. Bear has m workers in total, and the workers must work side by side. So at a time bear can choose some continuous houses, no more than m, and add their heights by one, this takes one day to finish.

Given the designed height for each house, what is the minimum number of days after which all the houses’ heights are no less than the original design?

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

#include <stack>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define test freopen("test.txt","r",stdin)

#define maxn 200000

#define mod 10007

#define eps 1e-9

int Num;

char CH[];

const int inf=0x3f3f3f3f;

const ll infll = 0x3f3f3f3f3f3f3f3fLL;

inline ll read()

{

    ll x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

inline void P(int x)

{

    Num=;if(!x){putchar('');puts("");return;}

    while(x>)CH[++Num]=x%,x/=;

    while(Num)putchar(CH[Num--]+);

    puts("");

}

//**************************************************************************************

ll dp[maxn];

void solve()

{

    ll n,m,a,sum=,ans=;

    n=read(),m=read();

    for(int i=;i<n;i++)

    {

        a=read();

        if(i>=m)sum-=dp[i-m];

        dp[i]=;

        if(a>sum)

        {

            ans+=a-sum;

            dp[i]=a-sum;

            sum=a;

        }

    }

    cout<<ans<<endl;

}

int main()

{

    //test;

    int t=read();

    for(int cas=;cas<=t;cas++)

    {

        printf("Case #%d: ",cas);

        solve();

    }

}