Problem:
Score : 800 points

Problem Statement

You are given a string S consisting of lowercase English letters.
Determine whether we can turn S into a palindrome by repeating the operation of swapping two adjacent characters. If it is possible, find the minimum required number of operations.

Constraints

1 \leq |S| \leq 2 × 10^5

S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If we cannot turn S into a palindrome, print -1. Otherwise, print the minimum required number of operations.

Sample Input 1

eel

Sample Output 1

1

We can turn S into a palindrome by the following operation:

Swap the 2-nd and 3-rd characters. S is now ele.

Sample Input 2

ataatmma

Sample Output 2

4

We can turn S into a palindrome by the following operation:

Swap the 5-th and 6-th characters. S is now ataamtma.

Swap the 4-th and 5-th characters. S is now atamatma.

Swap the 3-rd and 4-th characters. S is now atmaatma.

Swap the 2-nd and 3-rd characters. S is now amtaatma.

Sample Input 3

snuke

Sample Output 3

-1

We cannot turn S into a palindrome.

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

#include <algorithm>
#include <bitset>
#include <climits>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>

#define MOD 1000000007
#define MOD2 998244353
#define int long long
#define EPS 1e-9
//#define PI 3.14159265358979

#define rep(i, n) for (int i = 0; i < (int)(n); i++)

using namespace std;

template < typename T >
ostream &operator<<(ostream &os, const vector< T > &A) {
	for (int i = 0; i < A.size(); i++)
		os << A[i] << " ";
	os << endl;
	return os;
}
template <>
ostream &operator<<(ostream &os, const vector< vector< int > > &A) {
	int N = A.size();
	int M;
	if (N > 0)
		M = A[0].size();
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < M; j++)
			os << A[i][j] << " ";
		os << endl;
	}
	return os;
}

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

struct edge {
	int from, to, d, c;
	edge(int _from = 0, int _to = 0, int _d = 0, int _c = 0) {
		from = _from;
		to = _to;
		d = _d;
		c = _c;
	}
	bool operator<(const edge &rhs) const {
		return (d == rhs.d) ? (c < rhs.c) : (d < rhs.d);
	}
};
struct aabb {
	int x1, y1, x2, y2;
	aabb(int x1, int y1, int x2, int y2) : x1(x1), y1(y1), x2(x2), y2(y2) {}
};

typedef vector< edge > edges;
typedef vector< edges > graph;
struct flow {
	int to, cap, rev, cost;
	flow(int to = 0, int cap = 0, int rev = 0, int cost = 0) : to(to), cap(cap), rev(rev), cost(cost) {}
};
typedef vector< vector< flow > > flows;

const int di[4] = {0, -1, 0, 1};
const int dj[4] = {-1, 0, 1, 0};
const int ci[5] = {0, 0, -1, 0, 1};
const int cj[5] = {0, -1, 0, 1, 0};
const ll LINF = LLONG_MAX / 2;
const int INF = INT_MAX / 2;
const double PI = acos(-1);

int pow2(int n) { return 1LL << n; }
template < typename T, typename U >
bool chmin(T &x, const U &y) {
	if (x > y) {
		x = y;
		return true;
	}
	return false;
}
template < typename T, typename U >
bool chmax(T &x, const U &y) {
	if (x < y) {
		x = y;
		return true;
	}
	return false;
}

struct initializer {
	initializer() {
		cout << fixed << setprecision(20);
	}
};
initializer _____;

int N, M, K, T, Q, H, W;
signed main() {
	string S;
	cin >> S;
	N = S.size();
	vector< int > count(26, 0);
	rep(i, N) count[S[i] - 'a']++;
	int p = 0;
	rep(i, 26) if (count[i] & 1) {
		if (p == 0)
			++p;
		else {
			cout << -1 << endl;
			return 0;
		}
	}
	int ans = 0;
	rep(j, 26) {
		char c = j + 'a';
		vector< int > index;
		rep(i, N) if (S[i] == c) index.push_back(i);
		rep(i, count[j] / 2) {
			ans += abs(N - 1 - index[count[j] - 1 - i] - index[i]);
		}
		if (count[j] & 1) {
			ans += abs(index[count[j] / 2] - N / 2);
		}
	}
	cout << ans / 2 << endl;
	return 0;
}