A lot of people in Berland hates rain, but you do not. Rain pacifies, puts your thoughts in order. By these years you have developed a good tradition — when it rains, you go on the street and stay silent for a moment, contemplate all around you, enjoy freshness, think about big deeds you have to do. 

Today everything had changed quietly. You went on the street with a cup contained water, your favorite drink. In a moment when you were drinking a water you noticed that the process became quite long: the cup still contained water because of rain. You decided to make a formal model of what was happening and to find if it was possible to drink all water in that situation. 

Thus, your cup is a cylinder with diameter equals *d* centimeters. Initial level of water in cup equals *h* centimeters from the bottom. 

You drink a water with a speed equals *v* milliliters per second. But rain goes with such speed that if you do not drink a water from the cup, the level of water increases on *e* centimeters per second. The process of drinking water from the cup and the addition of rain to the cup goes evenly and continuously. 

Find the time needed to make the cup empty or find that it will never happen. It is guaranteed that if it is possible to drink all water, it will happen not later than after 10<sup class="upper-index">4</sup> seconds.

Note one milliliter equals to one cubic centimeter.
The only line of the input contains four integer numbers *d*, *h*, *v*, *e* (1 ≤ *d*, *h*, *v*, *e* ≤ 10<sup class="upper-index">4</sup>), where:
 -  *d* — the diameter of your cylindrical cup, -  *h* — the initial level of water in the cup, -  *v* — the speed of drinking process from the cup in milliliters per second, -  *e* — the growth of water because of rain if you do not drink from the cup.
In the first example the water fills the cup faster than you can drink from it.

In the second example area of the cup's bottom equals to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/419dc74dcd7bc392019c9fe748fe1fdb08ab521a.png" style="max-width: 100.0%;max-height: 100.0%;"/>, thus we can conclude that you decrease the level of water by <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/e8edb237e1f805fe83c2f47e48d3a9d03f2ee304.png" style="max-width: 100.0%;max-height: 100.0%;"/> centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup will be empty in <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9dae615d7e2c5c7c03cb478848fb06aba1a8942e.png" style="max-width: 100.0%;max-height: 100.0%;"/> seconds.
1 2 3 100
NO
1 1 1 1
YES
3.659792366325
48 7946 7992 72
NO
72 6791 8546 46
NO
100 5635 9099 23
NO
20 287 3845 5
YES
39.646277165210
48 6428 9807 83
NO
72 5272 4552 64
NO
100 4117 5106 34
NO
20 2961 9852 15
YES
180.991437129723
48 1805 3109 93
NO
72 8534 7042 65
NO
1 47 80 68
YES
1.388102806810
4 495 8813 1
YES
0.706823517575
5 2797 5925 9
YES
9.553973511669
1 8324 4362 23
YES
1.505007106354
6 1976 8455 3
YES
6.674898722265
7 2644 8080 5
YES
12.900417790197
3 4183 5491 98
YES
6.162185601824
2 8591 320 101
YES
9999.259991757254
10000 10000 10000 10000
NO
2 5000 12 3
YES
6099.653943875812
10 1000 100 1
YES
3659.792366325487
#include <bits/stdc++.h>
#define ll long long
#define PI acos(-1)
#define MOD int(1e9+7)
using namespace std;
ll GCD(ll a, ll b){ return(b == 0) ? a : GCD(b, a%b); }
ll LCM(ll a, ll  b){ return(a*b)/GCD(a,b);}
int main() {
  //Parparosa
  #ifndef ONLINE_JUDGE
  freopen("input.txt","r",stdin);
  freopen("output.txt","w",stdout);
  #endif
  double d,h,e,v;
  cin>>d>>h>>v>>e;
  d/=2.0;
  double in=PI*d*d*e;
  double all=PI*d*d*h;
  if(in>v)cout<<"NO"<<endl;
  else
  {
    printf("YES\n%.12f",all/(v-in));
  }
}