Problem: 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. 

<image>

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 104 seconds.

Note one milliliter equals to one cubic centimeter.

Input

The only line of the input contains four integer numbers d, h, v, e (1 ≤ d, h, v, e ≤ 104), 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. 

Output

If it is impossible to make the cup empty, print "NO" (without quotes).

Otherwise print "YES" (without quotes) in the first line. In the second line print a real number — time in seconds needed the cup will be empty. The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4. It is guaranteed that if the answer exists, it doesn't exceed 104.

Examples

Input

1 2 3 100


Output

NO


Input

1 1 1 1


Output

YES
3.659792366325

Note

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 <image>, thus we can conclude that you decrease the level of water by <image> centimeters per second. At the same time water level increases by 1 centimeter per second due to rain. Thus, cup will be empty in <image> seconds.

Solution (in JAVA):

Solution:
import java.util.*;
import java.lang.*;
import java.io.*;



/* Name of the class has to be "Main" only if the class is public. */



public class Main
{
     
     
     
    static PrintWriter out;
    
	static class FastReader{
		BufferedReader br;
		StringTokenizer st;
		public FastReader(){
			br=new BufferedReader(new InputStreamReader(System.in));
			out=new PrintWriter(System.out);
		}
		String next(){
			while(st==null || !st.hasMoreElements()){
				try{
					st= new StringTokenizer(br.readLine());
				}
				catch (IOException e){
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		int nextInt(){
			return Integer.parseInt(next());
		}
		long nextLong(){
			return Long.parseLong(next());
		}
		double nextDouble(){
			return Double.parseDouble(next());
		}
		String nextLine(){
			String str = "";
			try{
				str=br.readLine();
			}
			catch(IOException e){
				e.printStackTrace();
			}
			return str;
		}
	}
	

// SHIVAM GUPTA :



// ASCII = 48 + i ;

// SHIVAM GUPTA :





public static int  min(int a ,int b , int c, int d)
{
       
       
       int[] arr = new int[4] ;
       arr[0] = a;
       arr[1] = b ;
       arr[2] = c;
       arr[3] = d;
       Arrays.sort(arr) ;
       
       return arr[0];
       
}
 
public static int  max(int a ,int b , int c, int d)
{
       
       
       int[] arr = new int[4] ;
       arr[0] = a;
       arr[1] = b ;
       arr[2] = c;
       arr[3] = d;
       Arrays.sort(arr) ;
       
       return arr[3];
       
}
 
 
static int sieve =  1000000 ;

 
static boolean[] prime =  new boolean[sieve + 1] ;

 
public static void sieveOfEratosthenes() 
    { 
        // TRUE == prime
        
        // FALSE ==  COMPOSITE
        
        // FALSE== 1
       
        for(int i=0;i< sieve + 1;i++) 
            prime[i] = true; 
          
        for(int p = 2; p*p <= sieve; p++) 
        { 
           
            if(prime[p] == true) 
            { 
                
                for(int i = p*p; i <= sieve; i += p) 
                    prime[i] = false; 
            } 
        } 
          
       
       
    
    } 
 
 
public static String reverse(String input)
{
    String op = "" ;
    
    for(int i = 0; i < input.length() ; i++ )
    {
        op = input.charAt(i)+  op ; 
    }
    
return op ;
}




public static int[] sortI(int[] arr)
{
      Arrays.sort(arr) ;
      return arr ;
}

public static int[] sortD(int[] arr)
{
      Arrays.sort(arr) ;
      
      int i =0 ; int j = arr.length -1  ;
      
      while( i < j)
      {
            int temp = arr[i] ;
            arr[i] =arr[j] ;
            arr[j] = temp ;
            i++ ; j-- ;
      }
      
      
      
      return arr ;
}



public static boolean isPossibleTriangle(int a  ,int b , int c)
{
      if( a + b > c && c+b > a && a +c > b)
      {
            return true  ;
      }
      
      return false  ;
      
      
}
    
public static int gcd(int a, int b )
{

if(b==0)return a ;

else return gcd(b,a%b) ; 


}


public static int lcm(int a, int b ,int c , int d )
{

int temp = lcm(a,b , c) ;


 
 int ans = lcm(temp ,d ) ;

return ans  ;


}



public static int lcm(int a, int b ,int c )
{

int temp = lcm(a,b) ;

int ans =  lcm(temp ,c) ;

return ans  ;


}


    
public static int lcm(int a , int b )
{

int gc = gcd(a,b);

return (a*b)/gc ;




}

   
    public static int countDigit(long n) 
    { 
        return (int)Math.floor(Math.log10(n) + 1); 
    } 
    
  
  
   static final int MAXN = 100001; 
       
    
    static int spf[] = new int[MAXN]; 
   
    static void sieve() 
    { 
        spf[1] = 1; 
        for (int i=2; i<MAXN; i++) 
       
           
            spf[i] = i; 
       
       
        for (int i=4; i<MAXN; i+=2) 
            spf[i] = 2; 
       
        for (int i=3; i*i<MAXN; i++) 
        { 
           
            if (spf[i] == i) 
            { 
               
                for (int j=i*i; j<MAXN; j+=i) 
       
                    if (spf[j]==j) 
                        spf[j] = i; 
            } 
        } 
    } 
       
// The above code works well for n upto the order of 10^7.
// Beyond this we will face memory issues.

// Time Complexity: The precomputation for smallest prime factor is done in O(n log log n)
// using sieve.
// Where as in the calculation step we are dividing the number every time by 
// the smallest prime number till it becomes 1.
// So, let’s consider a worst case in which every time the SPF is 2 .
// Therefore will have log n division steps.


// Hence, We can say that our Time Complexity will be O(log n) in worst case.


    static Vector<Integer> getFactorization(int x) 
    { 
        Vector<Integer> ret = new Vector<>(); 
        while (x != 1) 
        { 
            ret.add(spf[x]); 
            x = x / spf[x]; 
        } 
        return ret; 
    } 
       
   
    
public static void main (String[] args) throws java.lang.Exception
{
  FastReader scn = new FastReader() ;

double d = scn.nextInt() ; double h = scn.nextInt() ;
double v = scn.nextInt() ; double e = scn.nextInt() ;
//int x3 = scn.nextInt() ; int y3 = scn.nextInt() ;


double rate = (4*v)/(Math.PI*d*d) ;

if( e >= rate )
{
      out.println("NO") ;
}

else{
      
      out.println("YES");
      out.println(h/(rate-e)) ;
      
      
}


out.flush() ;
 
  
}


}