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)

/*input
10
3 14 159 2653 58979 323846 2643383 27950288 419716939 9375105820

*/
 
/*
                                                    ______________
                                                          |       )
                                                          |   )   /
                                                          |______/
                                                    |     |      \
                                                    |     |   )   \
                                                    |_____|_______)
 
*/
 
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>
#include <iomanip>
#include <fstream>

using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll , ll> PLL;
typedef long double ld;
 
#define pb push_back
#define all(c) c.begin(),c.end()
#define allr(c) c.rbegin(),c.rend()
int mod = 1000000007;
const int inf = 1034567891;
const ll LL_INF = 1234567890123456789ll;
#define PI 3.14159265
#define endl '\n'
#define F first
#define S second
#define debug(x) cout << #x << " = " << x << endl;
#define TRACE
 
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
  cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
  const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif
 
#define out(container) for (auto it : container) cout << it << " "; cout << endl;
 
 
template < typename T > T GCD(T a, T b)            { ll t; while(a) { t = a; a = b % a; b = t; } return b; }
template < typename T > string toString(T a)       { return to_string(a); }
template < typename T > void toInt(string s, T &x) { stringstream str(s); str >> x;}
inline int add(int x, int y){ x += y; if(x >= mod) x -= mod; return x;}
inline int sub(int x, int y){ x -= y; if(x < 0) x += mod; return x;}
inline int mul(int x, int y){ return (x * 1ll * y) % mod;}
inline int powr(int a, ll b){
  int x = 1 % mod;
  while(b){
    if(b & 1) x = mul(x, a);
    a = mul(a, a);
    b >>= 1;
  }
  return x;
}
inline int inv(int a){ return powr(a, mod - 2);}

int main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  
  string str;
  cin >> str;
  int n = str.size();
  int ans = 0;
  for (int i = 0; i < n / 2; i++) {
    ans += (str[i] != str[n - i - 1]);
  }
  cout << ans << endl;


  return 0;
}