// Codeforces Round #365 (Div. 2)

 // C - Chris and Road 二分找切点

 // 题意：给你一个凸边行，凸边行有个初始的速度往左走，人有最大速度，可以停下来，竖直走。

 // 问走到终点的最短时间

 // 思路：

 // 1.贪心来做

 // 2.我觉的二分更直观

 // 可以抽象成：一条射线与凸边行相交，判断交点。二分找切点

 #include <bits/stdc++.h>

 using namespace std;

 #define LL long long

 const int inf = 0x3f3f3f3f;

 const int MOD =;

 const int N =1e6+;

 #define clc(a,b) memset(a,b,sizeof(a))

 const double eps = 1e-;

 struct Point{

     double x,y;

     Point(){}

     Point(int x_,int y_):x(x_),y(y_){}

     Point operator + (const Point &t)const{

         return Point(x+t.x,y+t.y);

     }

     Point operator - (const Point &t)const{

         return Point(x-t.x,y-t.y);

     }

     double operator * (const Point &t)const{

         return x*t.y-y*t.x;

     }

     double operator ^ (const Point &t)const{

         return x*t.x+y*t.y;

     }

     bool operator < (const Point & a) const{

         if(x==a.x) return y>a.y;

         return x<a.x;

     }

 }p[];

 int n;

 bool judge(double k,double b){

     int tot=,cnt=;

     for(int i=;i<=n;i++){

         if(k*p[i].x+b-p[i].y<) tot++;

         else cnt++;

         if(cnt&&tot) return false;

     }

     return true;

 }

 int main(){

     int w;

     double v,u;

     scanf("%d%d%lf%lf",&n,&w,&v,&u);

     for(int i=;i<=n;i++){

         scanf("%lf%lf",&p[i].x,&p[i].y);

     }

     double k=u/v;

     if(judge(k,0.0)){

         printf("%.8f\n",(double)(w/u));

     }

     else{

         double l=,r=1e9;

         double ans;

         while(r-l>eps){

              double mid = (l+r)/2.0;

              if(judge(k,-mid/v*u)){

                 ans=(double)(w/u)+mid/v;

                 r=mid;

              }

              else{

                 l=mid;

              }

         }

         printf("%.8f\n",ans);

     }

     return ;

 }