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)

#include <bits/stdc++.h>

#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define INF 999999999
#define size_t unsigned long long
#define ll long long
#define rep(i,a) for(int i=0;i<(a);i++)
#define repr(i,a) for(int i=(int)(a)-1;i>=0;i--)
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define FORR(i,a,b) for(int i=(int)(b)-1;i>=a;i--)
#define ALL(a) a.begin(), a.end()
using namespace std;
int si() { int x; scanf("%d", &x); return x; }
long long sl() { long long x; scanf("%lld", &x); return x; }
string ss() { string x; cin >> x; return x; }
void pi(int x) { printf("%d ", x); }
void pl(long long x) { printf("%lld ", x); }
void pd(double x) { printf("%.15f ", x); }
void ps(const string &s) { printf("%s ", s.c_str()); }
void br() { putchar('\n'); }

const int MOD = 1e9 + 7;

struct mint {
    int n;
    mint(int n_ = 0) : n(n_) {}
};

mint operator+(mint a, mint b) { a.n += b.n; if (a.n >= MOD) a.n -= MOD; return a; }
mint operator-(mint a, mint b) { a.n -= b.n; if (a.n < 0) a.n += MOD; return a; }
mint operator*(mint a, mint b) { return (long long)a.n * b.n % MOD; }
mint &operator+=(mint &a, mint b) { return a = a + b; }
mint &operator-=(mint &a, mint b) { return a = a - b; }
mint &operator*=(mint &a, mint b) { return a = a * b; }

typedef pair<int, int> P;

const int N = 1e6+5;
string s;
vector<int> v[N];
int n,c;
ll a[N];

int main () {
  cin >> s;
  n = s.length()/2;
  rep(i,n){
    if(s[i] != s[s.length()-1-i]) c++;
  }
  cout << c << endl;
}