Codeforces Beta Round #9 (Div. 2 Only)：B

B. Running Student

time limit per test
1 second

memory limit per test
64 megabytes

input
standard input

output
standard output

And again a misfortune fell on Poor Student. He is being late for an exam.

Having rushed to a bus stop that is in point (0, 0), he got on a minibus and they drove along a straight line, parallel to axis OX,
in the direction of increasing x.

Poor Student knows the following:

during one run the minibus makes n stops, the i-th
stop is in point (xi, 0)

coordinates of all the stops are different

the minibus drives at a constant speed, equal to vb

it can be assumed the passengers get on and off the minibus at a bus stop momentarily

Student can get off the minibus only at a bus stop

Student will have to get off the minibus at a terminal stop, if he does not get off earlier

the University, where the exam will be held, is in point (xu, yu)

Student can run from a bus stop to the University at a constant speed vs as
long as needed

a distance between two points can be calculated according to the following formula: 


Student is already on the minibus, so, he cannot get off at the first bus stop

Poor Student wants to get to the University as soon as possible. Help him to choose the bus stop, where he should get off. If such bus stops are multiple, choose the bus stop closest to the University.

Input

The first line contains three integer numbers: 2 ≤ n ≤ 100, 1 ≤ vb, vs ≤ 1000.
The second line contains n non-negative integers in ascending order: coordinates xi of
the bus stop with index i. It is guaranteed that x1 equals
to zero, and xn ≤ 105.
The third line contains the coordinates of the University, integers xu and yu,
not exceeding 105 in
absolute value.

Output

In the only line output the answer to the problem — index of the optimum bus stop.

Examples

input

4 5 2
0 2 4 6
4 1

output

3

input

2 1 1
0 100000
100000 100000

output

2

Note

As you know, students are a special sort of people, and minibuses usually do not hurry. That's why you should not be surprised, if Student's speed is higher than the speed of the minibus.

题意：一个男孩要去上学，现在有搭车水平行驶以及走路两种方案，或者先搭一段路车后走路。求到达学校所用时间最少。（现在男孩在公交车上，即在第一个站是不能下车的）。

题解：直接暴力枚举在某点下车所用时间，然后更新答案就可以了。

//Wud
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <sstream>
#include <fstream>
#include <set>
#include <map>
#include <string.h>
#define EPS 1e-9
#define M_PI 3.14159265358979323846
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
const int maxn = 5e6+7;
int n,vb,vs;
int s[150];
int x,y;
int main()
{
while(scanf("%d %d %d",&n,&vb,&vs)!=EOF)
{
for (int i = 0;i < n;i++) scanf("%d",&s[i]);
scanf("%d %d",&x,&y);
double ans = 1e18;
int p;
int clo;
for (int i = 1;i < n;i++)
{
double len = sqrt(1ll*(s[i]-x)*(s[i]-x)+1ll*y*y);
double t = s[i]*1.0/vb+len*1.0/vs;
if(t<ans){
ans = t;
p = i;
clo = abs(s[i]-x);
}
if(t==ans)
{
if(clo>abs(s[i]-x)){
p = i;
clo = abs(s[i]-x);
}
}
}
printf("%d\n",p+1);
}
return 0;
}