Today, Husam was visiting his grandmother. Husam always takes a path such that the total distance he will walk is as minimum as possible. Husam decided to give you all information about his journey so that you can find the total distance he walked.

Husam grandmother’s home is a kilometers away from Husam’s home to the south, and b kilometers away to the west. Also, there is a river located c kilometers away from Husam’s home to the east.

Before Husam visiting his grandmother, he went to the river to fill his bottle with water. After this, he then went directly to his grandmother’s home. After Husam finished his visit, he followed the direct straight path between his grandmother’s home and his home. When Husam walked x% of that path, he remembered that his bottle is empty, so, he went to the river to fill it, then he went to his home.

Husam needs your help to calculate the distance he walked today. Can you?

Input

The first line contains an integer T (1 ≤ T ≤ 10^5) specifying the number of test cases.

Each test case consists of a single line containing four integers a, b, c and x (1 ≤ a, b, c ≤ 10^9, 0 ≤ x ≤ 100).

Consider the ground and the river as a flat land, and the length of the river is limitless. Also, you can consider Husam and his grandmother homes as small points.
Output

For each test case, print a single line containing the total distance Husam walked during his journey.

Your answer will be considered correct if its relative or absolute error does not exceed 10 ^- 9.
Example
Input

1
6 2 3 18

Output

20.225553719

最短路
初中数学问题？

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int t;
double a,b,c,x,ans,d,e;
int main(){
	scanf("%d",&t);
	while(t--){
		scanf("%lf%lf%lf%lf",&a,&b,&c,&x);
		ans=sqrt((a*a)+(b+2*c)*(b+2*c));
		d=b/100*x;
		e=a/100*x;
		ans+=sqrt(d*d+e*e);
		ans+=sqrt((b-d+2*c)*(b-d+2*c)+(a-e)*(a-e));
		printf("%.9f\n",ans);
	}
	return 0;
}