FSF is addicted to a stupid tower defense game. The goal of tower defense games is to try to stop enemies from crossing a map by building traps to slow them down and towers which shoot at them as they pass.

The map is a line, which has n unit length. We can build only one tower on each unit length. The enemy takes t seconds on each unit length. And there are 3 kinds of tower in this game: The red tower, the green tower and the blue tower. 

The red tower damage on the enemy x points per second when he passes through the tower.

The green tower damage on the enemy y points per second after he passes through the tower.

The blue tower let the enemy go slower than before (that is, the enemy takes more z second to pass an unit length, also, after he passes through the tower.)

Of course, if you are already pass through m green towers, you should have got m*y damage per second. The same, if you are already pass through k blue towers, the enemy should have took t + k*z seconds every unit length.

FSF now wants to know the maximum damage the enemy can get.

There are multiply test cases.

The first line contains an integer T (T<=100), indicates the number of cases. 

Each test only contain 5 integers n, x, y, z, t (2<=n<=1500,0<=x, y, z<=60000,1<=t<=3)

For each case, you should output "Case #C: " first, where C indicates the case number and counts from 1. Then output the answer. For each test only one line which have one integer, the answer to this question.

1
2 4 3 2 1

Case #1: 12

Hint
For the first sample, the first tower is blue tower, and the second is red tower. So, the total damage is 4*(1+2)=12 damage points.
 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1505;
typedef __int64 LL;
LL dp[N][N];
int main()
{
    LL x, y, z, t, n, i, j;
    int T, cas = 0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t);
        memset(dp, 0, sizeof(dp));
        LL ans = n * t * x; //全部放红塔
        for(i = 1; i <= n; i++)  //前i个单位长度
        {
            for(j = 0; j <= i; j++) // 蓝塔数量
            {
                if(j == 0)
                    dp[i][j] = dp[i-1][j] + (i - j - 1) * y * t;
                else
                {
                    LL tmp1 = dp[i-1][j-1] + (i - j) * y * (t + (j - 1) * z); //第j个单位长度放蓝塔
                    LL tmp2 = dp[i-1][j] + (i - j - 1) * y * (t + j * z); //第j个单位长度不放蓝塔
                    dp[i][j] = max(tmp1, tmp2);
                }
                ans = max(ans, dp[i][j] + (n - i) * x * (t + j * z) + (n - i) * (t + j * z) * (i - j) * y);
            }
        }
        printf("Case #%d: %I64d\n", ++cas, ans);
    }
    return 0;
}