# Codeforces 937C--Save Energy!

xiaoxiao2021-03-01  5

Julia is going to cook a chicken in the kitchen of her dormitory. To save energy, the stove in the kitchen automatically turns off after k minutes after turning on.

During cooking, Julia goes to the kitchen every d minutes and turns on the stove if it is turned off. While the cooker is turned off, it stays warm. The stove switches on and off instantly.

It is known that the chicken needs t minutes to be cooked on the stove, if it is turned on, and 2t minutes, if it is turned off. You need to find out, how much time will Julia have to cook the chicken, if it is considered that the chicken is cooked evenly, with constant speed when the stove is turned on and at a constant speed when it is turned off.

Input

The single line contains three integers k, d and t (1 ≤ k, d, t ≤ 1018).

Output

Print a single number, the total time of cooking in minutes. The relative or absolute error must not exceed 10 - 9.

Examples Input Copy 3 2 6 Output 6.5 Input Copy 4 2 20 Output 20.0 Note

In the first example, the chicken will be cooked for 3 minutes on the turned on stove, after this it will be cooked for . Then the chicken will be cooked for one minute on a turned off stove, it will be cooked for . Thus, after four minutes the chicken will be cooked for . Before the fifth minute Julia will turn on the stove and after 2.5 minutes the chicken will be ready .

In the second example, when the stove is turned off, Julia will immediately turn it on, so the stove will always be turned on and the chicken will be cooked in 20 minutes.

一开始的时候 人按下按钮后就走了，然后过了一段时间后，当烤炉刚好保温的时候 人过来了，把烤炉按上，接着烤。

鸡需要t秒烤的热量--->t*=2;

烤炉烤的时候提供的热量--->k*2

烤炉保温的时候提供的热量--->t2

（1） 烤炉还在运行的时候 鸡就烤好了

（2）烤炉在保温的时候 鸡才好

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; long long k,d,t; // k 烤炉的时间 // d 人走的时间间隔 // t 所需时间 int main(){ scanf("%I64d%I64d%I64d",&k,&d,&t); t*=2; long long t1=k,t2=d-(k%d); if(t2==d) t2=0; long long xun=2*k+t2; double te=(t/xun); double ans=(1.0*te)*(t1+t2); t%=xun; //printf("%lld %lld %lld %lf",t1,t2,xun,ans); if(t<=2*k){ ans+=t/2.0; } else{ t-=2*k; ans+=k; ans+=t; } printf("%.1lf\n",ans); return 0; }