Pairs with Sum divisible by 5 You will be given two integers \$M\$ and \$N\$. Find the number of pairs \$(x,y)\$ such that \$1 \le x \le M\$, \$1 \le y \le N\$ and \$(x+y)\mod5 = 0\$.
For example, if \$M = 6\$ and \$N = 12\$, pairs which satisfies such conditions are, \$(1,4), (4,1), (1,9), (2,3), (2,8), (3,2), (3,7), (3,12), (4,6), (6,4), (4,11), (5,5), (5,10), (6,9)\$
Total \$14\$.
Sample
Input : 6 12
Output: 14

Input : 11 14
Output: 31

Input : 553 29
Output: 3208

Input : 2 2
Output: 0

Input : 752486 871672
Output: 131184195318 

This is a code-golf challenge so code with lowest bytes wins!
Update
Jonathan Allan's solution has the smallest code size, 5 bytes. However, it doesn't produce an answer for the last given test.
I have decided to go with the next answer with the shortest size that produces the correct answer for the largest test, there is a tie between two golfers who competed neck-to-neck.
I proudly present the winners of this challenge Lyxal and Kevin Cruijssen with only 7 bytes code! Congratulations! 
As many of you, I found Arnauld's answer most helpful in finding the correct solution. So, I am accepting Arnauld's answer.
Thank you, Golfers!
Python 2, 32 bytes



lambda m,n:(m*n+abs(m%5-~n%5))/5


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Adapts an idea from Arnauld. We estimate that \$1/5\$ of the \$mn\$ pairs will satisfy the mod-5 condition. Th actual value may be one higher depending on the mod-5 values of \$m\$ and \$n\$, and we bump up \$mn\$ a bit so that these make it pass the next multiple of 5.



34 bytes



lambda m,n:m*n/5+(0&lt;-m%5*(-n%5)&lt;5)


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34 bytes



lambda m,n:m*n/5+(m*n%5+m%5+n%5)/9


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36 bytes



lambda m,n:m*n/5+((5-m%5)*(5-n%5)&lt;5)


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36 bytes



lambda m,n:m*n/5+(abs(m%5+n%5*1j)&gt;4)


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36 bytes



lambda m,n:m*n/5+(m%5*5/4+n%5*5/4&gt;5)


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36 bytes



lambda m,n:m*n/5+(m%5+n%5+m*n%5/4&gt;5)


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