After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers — the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.

Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and <image> equals 0.

As usual, Alyona has some troubles and asks you to help.

Input

The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).

Output

Print the only integer — the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.

Examples

Input

6 12


Output

14


Input

11 14


Output

31


Input

1 5


Output

1


Input

3 8


Output

5


Input

5 7


Output

7


Input

21 21


Output

88

Note

Following pairs are suitable in the first sample case: 

  * for x = 1 fits y equal to 4 or 9; 
  * for x = 2 fits y equal to 3 or 8; 
  * for x = 3 fits y equal to 2, 7 or 12; 
  * for x = 4 fits y equal to 1, 6 or 11; 
  * for x = 5 fits y equal to 5 or 10; 
  * for x = 6 fits y equal to 4 or 9. 



Only the pair (1, 4) is suitable in the third sample case.
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
import java.lang.*;
import java.util.ArrayList;
import java.math.BigInteger;
public class me {

public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner br = new Scanner(System.in);
        int n=br.nextInt();
        int m=br.nextInt();
        long sum=0;
        for(int i=1;i<=n;i++)
        {
            int a=i%5;
            sum+=(long)((m+a)/5);
            //System.out.println(a+" "+sum);
        }System.out.print(sum);
       
    
}

     
     

                 /*palindrone number check*/
                 
int Palindrome(String s) {
  int n = s.length();
  for (int i = 0; i < (n/2); ++i) {
     if (s.charAt(i) != s.charAt(n - i - 1)) {
         return 0;
     }
  }

  return 1;
}
    public static long  ways(int n, int r, long MOD)
    {
      //  vector<long long> f(n + 1,1);
        long f[]=new long[n+1];
        Arrays.fill(f,1);
        for (int i=2; i<=n;i++)
            f[i]= (f[i-1]*i) % MOD;
        return (f[n]*((InverseEuler(f[r], MOD) * InverseEuler(f[n-r], MOD)) % MOD)) % MOD;
    }  
    
                  
                  
                  
                   /*THIS IS FOR THE INVERSE EULER*/
    public static long  InverseEuler(long n, long MOD)
    {
        return pow(n,MOD-2,MOD);
    }
    
    
    
                   
                   /*THIS ID THE CHECK FOR THE PRIME*/
    static int isPrime(int n) {
    for(int i=2;i<n;i++) {
        if(n%i==0)
            return 0;
    }
    return 1;
}
    
    
    
                          /*TO CHECK FOR THE GCD IN LONG*/
public static long gcd(long x,long y)
{
    if(x%y==0)
        return y;
    else
        return gcd(y,x%y);
}
    
    
    
                             /* CHECK FOR THE GCD IN INT*/
public static int gcd(int x,int y)
{
    if(x%y==0)
        return y;
    else 
        return gcd(y,x%y);
}


                         /*CHECK FOR THE ABSOLUTE IN INT*/
public static int abs(int a,int b)
{
    return (int)Math.abs(a-b);
}



                        /*CHECK FOR THE ABSOLUTE IN LONG*/
public static long abs(long a,long b)
{
    return (long)Math.abs(a-b);
}


                       /*CHECK FOR MAX BETWEEN TWO NUMBERS IN INT*/
public static int max(int a,int b)
{
    if(a>b)
        return a;
    else
        return b;
}



                       /*CHECK FOR MIN BETWEEN TWO NUMBERS IN INT*/
public static int min(int a,int b)
{
    if(a>b)
        return b;
    else 
        return a;
}


                       /*CHECK FOR MAX BETWEEN TWO NUMBERS IN LONG*/
public static long max(long a,long b)
{
    if(a>b)
        return a;
    else
        return b;
}

                     /*CHECK FOR MIN BETWEEN TWO NUMBERS IN LONG*/
public static long min(long a,long b)
{
    if(a>b)
        return b;
    else 
        return a;
}
                    /*check for power of two of any num*/
static boolean isPowerOfTwo (long v) {
    return (v & (v - 1)) == 0;
}
public static long pow(long n,long p,long m)
{
     long  result = 1;
      if(p==0)
        return 1;
    if (p==1)
        return n;
    while(p!=0)
    {
        if(p%2==1)
            result *= n;
        if(result>=m)
        result%=m;
        p >>=1;
        n*=n;
        if(n>=m)
        n%=m;
    }
    return result;
}

                    /*SORT GIVEN ARRAY*/
static long sort(int a[],int n)
{   int b[]=new int[n]; 
    return mergeSort(a,b,0,n-1);}


static long mergeSort(int a[],int b[],long left,long right)
{ long c=0;if(left<right)
 {   long mid=left+(right-left)/2;
     c= mergeSort(a,b,left,mid);
     c+=mergeSort(a,b,mid+1,right);
     c+=merge(a,b,left,mid+1,right); }  
    return c;    }


static long merge(int a[],int  b[],long left,long mid,long right)
{long c=0;int i=(int)left;int j=(int)mid; int k=(int)left;
while(i<=(int)mid-1&&j<=(int)right)
{ if(a[i]<=a[j]) {b[k++]=a[i++]; }
else    { b[k++]=a[j++];c+=mid-i;}}
while (i <= (int)mid - 1)   b[k++] = a[i++]; 
while (j <= (int)right) b[k++] = a[j++];
for (i=(int)left; i <= (int)right; i++) 
    a[i] = b[i];  return c;  }


    
    
                       /*REMOVE ZERO FROM GIVEN NUM*/
static int removeZero(int num) {
        int ret = 0;
        int ten = 1;
        while (num>0) {
                int dig = num % 10;
                num /= 10;
                if (dig>0) {
                        ret += dig * ten;
                        ten *= 10;
                }
        }
        return ret;
}
                 /*CHECK FOR LCM*/
 static int lcm(int x, int y)
    {
        int a;
        a = (x > y) ? x : y; // a is greater number
        while(true)
        {
            if(a % x == 0 && a % y == 0)
                return a;
            ++a;
        }	
    }
}