Problem: Captain Bill the Hummingbird and his crew recieved an interesting challenge offer. Some stranger gave them a map, potion of teleportation and said that only this potion might help them to reach the treasure. 

Bottle with potion has two values x and y written on it. These values define four moves which can be performed using the potion:

  * <image>
  * <image>
  * <image>
  * <image>



Map shows that the position of Captain Bill the Hummingbird is (x1, y1) and the position of the treasure is (x2, y2).

You task is to tell Captain Bill the Hummingbird whether he should accept this challenge or decline. If it is possible for Captain to reach the treasure using the potion then output "YES", otherwise "NO" (without quotes).

The potion can be used infinite amount of times.

Input

The first line contains four integer numbers x1, y1, x2, y2 ( - 105 ≤ x1, y1, x2, y2 ≤ 105) — positions of Captain Bill the Hummingbird and treasure respectively.

The second line contains two integer numbers x, y (1 ≤ x, y ≤ 105) — values on the potion bottle.

Output

Print "YES" if it is possible for Captain to reach the treasure using the potion, otherwise print "NO" (without quotes).

Examples

Input

0 0 0 6
2 3


Output

YES


Input

1 1 3 6
1 5


Output

NO

Note

In the first example there exists such sequence of moves:

  1. <image> — the first type of move 
  2. <image> — the third type of move 

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 ;




}


static boolean isPrime(int n)
{
      if(n==1)
      {
            return false  ;
      }
      
      boolean ans =  true  ;
      
      for(int i = 2; i*i <= n ;i++)
      {
            if(n% i ==0)
            {
                  ans = false  ;break ;
            }
      }
      
      
      return ans  ;
      
      
}

static boolean isPowerOfTwo(int n) 
{ 
    if(n==0) 
    return false; 
  
if(((n ) & (n-1)) == 0 ) return true ;
else return false  ;

} 
   
   
   
    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)) ;
      
      
// }

// int t = scn.nextInt() ;


// for(int i1 =1 ; i1<= t ; i1++)
// {
      
//       int n= scn.nextInt() ;
      
//     //"Ashishgup" if he wins, and "FastestFinger"
    
//     if( n==1 ) 
//     {
//     out.println("FastestFinger") ;
    
//     }
    
//     else if( n% 2 != 0 )
//     {
//           out.println("Ashishgup") ;
//     }
//     else{
          
//           if(n==2 )
//           {
//           out.println("Ashishgup") ;
//           }
          
//          // NO ODD DIVISOR 
//          // ONLY EVEN DIVISORS
         
//           else if(isPowerOfTwo(n))
//           {
//           out.println("FastestFinger") ;
//           }

     
//     // HAS ODD DIVISORS
//     // DIVIDE BY MAX ODD DIVISOR
//     // AFTER U WILL GET SOMETHING POWER OF 2 
//     // IF THAT NUMBER WHICH LEFT AT END
//     // IS 2 THEN FF WIN
//     // ELSE AGP WIN
     

//        else{
//              // n is divisible by 4
//              //AGP WILL DIIVIDE N BY LARGEST ODD DIVISOR
//              // THEN FF WILL GET ONLY NUM 2^X TYPE
//              // AND CAN ONLY SUBTRACT 1 AND GIVE BACK TO 
//              // AGP
             
//              if(n% 4 == 0)
//              {
//                    out.println("Ashishgup") ;
//              }
             
//              else{
//                    int k = n/2  ;
                   
//                 if(isPrime(k))
//                 {
//                       out.println("FastestFinger") ;
                      
//                 }
//                 else{
//                       out.println("Ashishgup") ;
//                 }
                   
           
//              }
             
//        }   
          
//     }
    
     
// }

String ans  ; 
 int x1  = scn.nextInt() ; int y1 = scn.nextInt() ;

int x2 = scn.nextInt() ; int y2 = scn.nextInt() ;

 int x  = scn.nextInt() ; int y = scn.nextInt() ;

 int modx =  (int)Math.abs(x1-x2) ;
 int mody =  (int)Math.abs(y1-y2) ;

if(modx % x !=0 || mody % y != 0 )
{
      ans = "NO" ;
}

else 
{
  int temp1 =  modx/x   ;
  int temp2 =  mody/y  ;
      
      if(temp1 % 2 == 0 && temp2 %2 ==0)
      {
        ans = "YES" ;    
      }
      else if(temp1 % 2 != 0 && temp2 %2 !=0)
      {
           ans = "YES" ;     
      }
      else{
            ans = "NO" ; 
      }
      
}

out.println(ans) ;

out.flush() ;
 
  
}


}