Codeforces Round #471 (Div. 2)A. Feed the cat

After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.

At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.

Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.

Input

The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.

The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).

Output

Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if .

Examples

input

19 00
255 1 100 1

output

25200.0000

input

17 41
1000 6 15 11

output

1365.0000

题意：给定Andrew醒来时间，醒来时候猫的饥饿度，猫每分钟增长饥饿度，一份猫食价格，一份猫食能消除的饥饿度；现在有两种方案，一种是醒来直接买猫食，一种是到20:00后买猫食可以打八折，输出最省钱方案花费的钱数。
复制粘贴出错了还过了预判，然后终判wa了，难受

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
#define eps 0.00000001
#define pn printf("\n")
using namespace std;
typedef long long ll;

int main()
{
    ll hh,mm,h,d,c,n;
    cin >> hh >> mm >> h >> d >> c >> n;
    if(hh < )
    {
        double cur = (h / n + (h%n != ))*1.0 * c;
        double cut = ((( *   - hh* - mm)*d + h)/n + ((( *   - hh* - mm)*d+h)%n != )) * c * 0.8;
        printf("%.4f\n",min(cur,cut));
    }
    else
        printf("%.4f\n", (h / n + (h%n != ))*0.8 * c);

}