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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(unused_imports)]

use text_io::*;
use proconio::*;
use std::collections::*;
use itertools::Itertools;
use std::process::exit;
use std::cmp::*;
use num::*;
use num::integer::Roots;

fn main() {
    input! {
    a:usize,
    b:String,
    c:String,
    d:String,
    }
    let mut b=b.chars().collect_vec();
    let mut c=c.chars().collect_vec();
    let mut d=d.chars().collect_vec();
    let mut e=0;
    for i in 0..b.len(){
        if b[i]==c[i]&&c[i]==d[i]{e+=0}
        else if b[i]==c[i]||c[i]==d[i]||d[i]==b[i]{e+=1}
        else{e+=2}
    }
    println!("{}",e)
}