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>

using namespace std;

typedef long long int ll;
typedef pair<int,int> P;

#define rep(i,n) for(int i=0;i<(n);++i)
#define repi(i,a,b) for(int i=int(a);i<(b);i++)
#define repr(i,b,a) for(int i=int(b);i>=(a);i--)
#define all(x) x.begin(),x.end()

const ll mod = 1e9+7;
const ll INF = 1e9;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
bool valid(int lx,int ux,int ly,int uy,int x,int y){
	return lx<=x&&x<ux&&ly<=y&&y<uy; 
}
ll power(ll x,ll p){
	if(p==0)return 1;
	ll res=power(x*x%mod,p/2);
	if(p%2==1)res=res*x%mod;
	return res;
}

int main()
{
	string s;
	cin>>s;
	int left=0,right=s.size()-1;
	int ans=0;
	while(left<=right){
		if(s[left]!=s[right])ans++;
		left++;
		right--;
	}
	cout<<ans<<endl;
	return 0;
}