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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <array>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <sstream>
#include <stdio.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <list>
#include <numeric>
#include <stack>
#include <iomanip>

//using namespace std;

#define Rep(i,a,b) for(int i = a; i < b; ++i)
#define rep(i,b) Rep(i,0,b)
#define allof(a) (a).begin(), (a).end()

typedef long long ll;
const int inf = 1e9 + 7;
const ll infll = 1ll << 60ll;
const ll mod = 1e9 + 7;
// 0~3までは右左下上 4~7までは斜め
constexpr int dx[] = { 1, 0, -1, 0, 1, 1, -1, -1 };
constexpr int dy[] = { 0, -1, 0, 1, 1, -1, -1, 1 };

template<typename T> void chmax(T& a, T b) { a = std::max(a, b); }
template<typename T> void chmin(T& a, T b) { a = std::min(a, b); }
template<typename T> void chadd(T& a, T b) { a = a + b; }

namespace {  // 名前なし名前空間
	// gcd, lcm は C++17 でstdに標準入りしたので言語のバージョンに注意
	// 最大公約数
	ll gcd(ll a, ll b) {
		if (b == 0) return a;
		return gcd(b, a % b);
	}

	// 最小公倍数
	ll lcm(ll a, ll b) {
		return a / gcd(a, b) * b;
	}

}

int main() {

	std::string s;
	std::cin >> s;

	int n = (int)s.size();

	int ans = 0;

	if (n % 2 == 1) {
		int half = n / 2;

		std::vector<char> v(half);
		rep(i, half) {
			v[i] = s[i];
		}
		for (int i = 0; i < half; ++i) {
			if (v[i] != s[n - 1 - i]) ++ans;
		}
	}
	else {
		int half = n / 2;

		std::vector<char> v(half);
		rep(i, half) {
			v[i] = s[i];
		}
		for (int i = 0; i < half; ++i) {
			if (v[i] != s[n - 1 - i]) ++ans;
		}
	}

	std::cout << ans << std::endl;

	return 0;
}