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.
from math import *
from copy import *
from string import *				# alpha = ascii_lowercase
from random import *
from sys import stdin
from sys import maxsize
from operator import *				# d = sorted(d.items(), key=itemgetter(1))
from itertools import *
from collections import Counter		# d = dict(Counter(l))
import math

def bin1(l,r,k,t,b,val,ans):
	if(l>r):
		return ans
	else:
		mid=(l+r)//2
		v=k**mid
		if(v==val):
			return v
		elif(v>val):
			ans=mid
			return bin1(mid+1,r,k,t,b,val,ans)
		else:
			return bin1(l,mid-1,k,t,b,val,ans)
		
def bin2(l,r,k,t,b,val,ans):
	if(l>r):
		return ans
	else:
		mid=(l+r)//2
		v=t*(k**mid)+b*(mid)
		if(v==val):
			return v
		elif(v>val):
			ans=mid
			return bin2(l,mid-1,k,t,b,val,ans)
		else:
			return bin2(mid+1,r,k,t,b,val,ans)

def SieveOfEratosthenes(n): 
      
    # Create a boolean array "prime[0..n]" and initialize 
    #  all entries it as true. A value in prime[i] will 
    # finally be false if i is Not a prime, else true. 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
          
        # If prime[p] is not changed, then it is a prime 
        if (prime[p] == True): 
              
            # Update all multiples of p 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    l=[]
    for i in range(2,n+1):
    	if(prime[i]):
    		l.append(i)
    return l
def bin(l,r,ll,val):
	if(l>r):
		return -1
	else:
		mid=(l+r)//2
		if(val>=ll[mid][0] and val<=ll[mid][1]):
			return mid
		elif(val<ll[mid][0]):
			return bin(l,mid-1,ll,val)
		else:
			return bin(mid+1,r,ll,val)
def deci(n):
	s=""
	while(n!=0):
		if(n%2==0):
			n=n//2
			s="0"+s
		else:
			n=n//2
			s="1"+s
	return s
def diff(s1,s2):
	if(len(s1)<len(s2)):
		v=len(s1)
		while(v!=len(s2)):
			s1="0"+s1
			v=v+1
	else:
		v=len(s2)
		while(v!=len(s1)):
			s2="0"+s2
			v=v+1
	c=0
	for i in range(len(s1)):
		if(s1[i:i+1]!=s2[i:i+1]):
			c=c+1
	return c
from sys import stdin, stdout 
def fac(a,b):
	v=a
	while(a!=b):
		v*=a-1
		a=a-1
	return v
def bino(l,r,n):
	if(l>r):
		return -1
	else:
		mid=(l+r)//2
		val1=math.log((n/mid)+1,2)
		val2=int(val1)
		if(val1==val2):
			return val1
		elif(val1<1.0):
			return bino(l,mid-1,n)
		else:
			return bino(mid+1,r,n)

def binary(l,r,ll,val,ans):
	if(l>r):
		return ans
	else:
		mid=(l+r)//2
		if(val%ll[mid]==0):
			ans=ll[mid]
			return binary(l,mid-1,ll,val,ans)
		else:
			return binary(l,mid-1,ll,val,ans)
def odd(n,ans,s):
	if(n%2!=0 or n in s):
		return ans
	else:
		s.add(n)
		return odd(n//2,ans+1,s)

if __name__ == '__main__':
	n,m=map(int,input().split(" "))
	count=0
	for i in range(1,n+1):
		val=(i%5)
		count+=((m+val)//5)
	print(count)