N = gets.to_i
a = gets
b = gets
c = gets

counter = 0

N.times{|i|
if a[i] != b[i] && b[i] != c[i] && c[i] != a[i]
    counter += 2
elsif a[i] == b[i] && b[i] == c[i]
    counter += 0
else
    counter += 1
end
}

puts counter
