Problem:
Score : 200 points

Problem Statement

Takahashi loves palindromes. Non-palindromic strings are unacceptable to him. Each time he hugs a string, he can change one of its characters to any character of his choice.

Given is a string S. Find the minimum number of hugs needed to make S palindromic.

Constraints

S is a string consisting of lowercase English letters.

The length of S is between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

S

Output

Print the minimum number of hugs needed to make S palindromic.

Sample Input 1

redcoder

Sample Output 1

1

For example, we can change the fourth character to o and get a palindrome redooder.

Sample Input 2

vvvvvv

Sample Output 2

0

We might need no hugs at all.

Sample Input 3

abcdabc

Sample Output 3

2

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
  ∧,,∧
( 'ω' )つ ＜WA,またお前か！！　
（m9 ＼ 　　
　 ＼　 ＼
　 　 ) ) ＼
　 ／／ ＼ ＼
　 (＿） 　 (＿)
*/

#include <bits/stdc++.h>
using namespace std;

#define all(x) (x).begin(),(x).end()
#define rep(x, y) for (int x = 0; x < y; x++)
#define MOD 1000000007

typedef long long LL;
typedef long double LD;

bool compare_by_b(pair<int, int> a, pair<int, int> b)
{
    if (a.second != b.second) {
        return a.second < b.second;
    } else {
        return a.first < b.first;
    }
}

ostringstream oss_global;
string s_global = oss_global.str();

int main()
{
    string s;
    cin >> s;

    if (s.length() == 1) {
        cout << "0" << endl;
        return 0;
    }

    int ans = 0;
    for (int i = 0; i <= s.length()/2 - 1; i++) {
        char a, b;
        a = s[i]; b = s[s.length() - i - 1];

        if (a - b) ans++;
    }

    cout << ans << endl;
    return 0;
}