Problem:
Score : 200 points

Problem Statement

You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters.

Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation:

Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter.

What is the minimum number of operations required to achieve the objective?

Constraints

1 \leq N \leq 100

Each of the strings A, B and C is a string of length N.

Each character in each of the strings A, B and C is a lowercase English letter.

Input

Input is given from Standard Input in the following format:

N
A
B
C

Output

Print the minimum number of operations required.

Sample Input 1

4
west
east
wait

Sample Output 1

3

In this sample, initially A = west、B = east、C = wait. We can achieve the objective in the minimum number of operations by performing three operations as follows:

Change the second character in A to a. A is now wast.

Change the first character in B to w. B is now wast.

Change the third character in C to s. C is now wast.

Sample Input 2

9
different
different
different

Sample Output 2

0

If A, B and C are already equal in the beginning, the number of operations required is 0.

Sample Input 3

7
zenkoku
touitsu
program

Sample Output 3

13

Solution in Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val n = readLine()!!.toInt()
    val a = readLine()!!
    val b = readLine()!!
    val c = readLine()!!
    val count = checkDiff(a, b, c)

    print(count)
}

fun checkDiff(a: String, b: String, c: String): Int {
    var count = 0

    a.forEachIndexed { index: Int, char: Char ->
        val matchAB = char == b[index]
        val matchAC = char == c[index]
        val matchBC = b[index] == c[index]

        val abc = matchAB && matchAC && matchBC
        val ab = matchAB && !matchAC && !matchBC
        val ac = !matchAB && !matchAC && matchBC
        val bc = !matchAB && matchAC && !matchBC

        count += when {
            abc -> 0
            ab -> 1
            ac -> 1
            bc -> 1
            else -> 2
        }
    }

    return count
}