#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>#include <vector>#include <utility>#include <stack>#include <queue>#include <map>#include <deque>#define max(x,y) ((x)>(y)?(x):(y))#define min(x,y) ((x)<(y)?(x):(y))using namespace std;long long f[1506][1506];long long ans,t,n,m,x,y,z;int main(){    int tt;    freopen("1005.in","r",stdin);    scanf("%d",&tt);    for(int ttt=1; ttt<=tt; ttt++)    {        scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t);        //无需memset(f) 只需保证递推关系完全依靠起始条件推出即可        ans=n*t*x;        f[0][0]=0;        for(int i=1; i<=n; i++)            for(int j=0; j<=i; j++)            {                if(j==0)                    f[i][j]=0;                else                {                    if(i-1>=j) f[i][j]=max(f[i-1][j]+j*y*(t+(i-1-j)*z),f[i-1][j-1]+(j-1)*y*(t+(i-j)*z));                    else f[i][j]=f[i-1][j-1]+(j-1)*y*(t+(i-j)*z);
　　　　　　　　　　　　//f[i][j]=max(f[i][j], f[i-1][j-1]+(j-1)*y*(t+(i-j)*z)) 不可 可能f[i][j]保存之前case的值                }                ans=max(ans, f[i][j]+(n-i)*(t+(i-j)*z)*x+(n-i)*(t+(i-j)*z)*j*y);            }        printf("Case #%d: %I64d\n",ttt,ans);    }    return 0;}