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 <iostream>
#include <iomanip>
#include <string>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <stack>
#include <queue>
#include <bitset>
#include <numeric>
#include <cassert>
#include <random>
#ifdef DEBUG
#include "./Lib/debug.hpp"
#include "./Lib/Timer.h"
#else
#define dump(...)
template<class T>inline auto d_val(T a, T b) { return a; }
#endif

/* (=＾o＾=) */
#define int ll

/* macro */
#define FOR(i, b, e) for(ll i = (ll)(b); i < (ll)(e); ++i)
#define RFOR(i, b, e) for(ll i = (ll)(e-1); i >= (ll)(b); --i)
#define REP(i, n) FOR(i, 0, n)
#define RREP(i, n) RFOR(i, 0, n)
#define REPC(x,c) for(const auto& x:(c))
#define REPI2(it,b,e) for(auto it = (b); it != (e); ++it)
#define REPI(it,c) REPI2(it, (c).begin(), (c).end())
#define RREPI(it,c) REPI2(it, (c).rbegin(), (c).rend())
#define REPI_ERACE2(it, b, e) for(auto it = (b); it != (e);)
#define REPI_ERACE(it, c) REPI_ERACE2(it, (c).begin(), (c).end())
#define ALL(x) (x).begin(),(x).end()
#define cauto const auto&
/* macro func */
#define SORT(x) do{sort(ALL(x));}while(false)
#define RSORT(x) do{sort((x).rbegin(),(x).rend());}while(false)
#define UNIQUE(v) do{v.erase( unique(v.begin(), v.end()), v.end() );}while(false)
#define MAX(x,y) do{x = std::max(x,y);}while(false)
#define MIN(x,y) do{x = std::min(x,y);}while(false)
#define BR do{cout<<"\n";}while(false)

/* type define */
using ll = long long;
using PAIR = std::pair<ll, ll>;
using VS = std::vector<std::string>;
using VL = std::vector<long long>;
using VVL = std::vector<VL>;
using VVVL = std::vector<VVL>;
using VD = std::vector<double>;
template<class T>
using V = std::vector<T>;

/* using std */
using std::cout;
constexpr char endl = '\n';
using std::cin;
using std::sort;
using std::pair;
using std::string;
using std::stack;
using std::queue;
using std::vector;
using std::list;
using std::map;
using std::unordered_map;
using std::multimap;
using std::unordered_multimap;
using std::set;
using std::unordered_set;
using std::unordered_multiset;
using std::multiset;
using std::bitset;
using std::priority_queue;

/* constant value */
constexpr ll MOD = 1000000007;
//constexpr ll MOD = 998244353;

/* Initial processing  */
struct Preprocessing { Preprocessing() { std::cin.tie(0); std::ios::sync_with_stdio(0); }; }_Preprocessing;

/* Remove the source of the bug */
signed pow(signed, signed) { assert(false); return -1; }

/* define hash */
namespace std { template <>	class hash<std::pair<ll, ll>> { public:	size_t operator()(const std::pair<ll, ll>& x) const { return hash<ll>()(1000000000 * x.first + x.second); } }; }

/* input */
template<class T> std::istream& operator >> (std::istream& is, vector<T>& vec) { for (T& x : vec) is >> x; return is; }

//=============================================================================================
/**
 *	遅延セグメント木を構成する
 *	4methodの変更により調整
 */
class LazySegmentTree {
private:
	const ll initialValue = 0;
	const ll ignoreValue = 0;
	const ll lazyIgnoreValue = 1e10;

	const ll m_size;
	std::vector<ll> m_node;
	std::vector<ll> m_lazy;

	ll calcSize(ll n) {
		ll size = 1;
		while (size < n) {
			size *= 2;
		}
		return size;
	}

	void spreadLazy(ll k) {
		if (k >= m_size - 1) { return; }
		m_lazy[2 * k + 1] = m_lazy[k] / 2;
		m_lazy[2 * k + 2] = m_lazy[k] / 2;
	}
	void reflectLazy(ll k) {
		m_node[k] += m_lazy[k];
	}

	ll merge(ll xl, ll xr) {
		return xl + xr;
	}
	ll rangeUpdate(ll x, ll l, ll r) {
		return x * (r - l);
	}
public:
	LazySegmentTree() = delete;
	LazySegmentTree(ll n) :
		m_size(calcSize(n)),
		m_node(m_size * 2 - 1, initialValue),
		m_lazy(m_size * 2 - 1, lazyIgnoreValue) {}

	void add(ll a, ll b, ll x) {
		add(a, b + 1, x, 0, 0, m_size);
		/**
		cout << "-- tree -- " << endl;
		REP(i, 2 * m_size - 1) {
			cout << m_node[i] << " " << m_lazy[i] << endl;
		}
		/*//**/
	}
	void add(ll a, ll b, ll x, ll k, ll l, ll r) {
		eval(k);
		if (r <= a || b <= l) { return; }
		if (a <= l && r <= b) {
			m_lazy[k] = rangeUpdate(x, l, r);
			eval(k);
			return;
		}
		add(a, b, x, 2 * k + 1, l, (l + r) / 2);
		add(a, b, x, 2 * k + 2, (l + r) / 2, r);
		m_node[k] = merge(m_node[2 * k + 1], m_node[2 * k + 2]);
	}
	void eval(ll k) {
		if (m_lazy[k] == lazyIgnoreValue) { return; }
		reflectLazy(k);
		spreadLazy(k);
		m_lazy[k] = lazyIgnoreValue;
	}
	ll query(ll a, ll b) { return query(a, b + 1, 0, 0, m_size); }
	ll query(ll a, ll b, ll k, ll l, ll r) {
		if (r <= a || b <= l) { return ignoreValue; }
		eval(k);
		if (a <= l && r <= b) { return m_node[k]; }
		return merge(
			query(a, b, 2 * k + 1, l, (l + r) / 2),
			query(a, b, 2 * k + 2, (l + r) / 2, r)
		);
	}
};

signed main() {
	string str;
	cin >> str;
	ll size = str.size();

	V<list<ll>> cv(26);
	REP(i, size) { cv[str[i] - 'a'].emplace_back(i); }

	ll key = -1;
	ll odd = 0;
	REP(i, 26) {
		if (cv[i].size() & 1) { ++odd; key = *std::next(cv[i].begin(), static_cast<ll>(cv[i].size() / 2)); }
	}
	if (odd > (0 + size & 1)) { cout << -1 << endl; return 0; }
	if (key > -1 && key < size / 2) {
		std::reverse(ALL(str));
		cv = V<list<ll >>(26);
		REP(i, size) { cv[str[i] - 'a'].emplace_back(i); }
	}

	auto tree = LazySegmentTree(size);

	VL lmt; lmt.reserve(26);
	REPC(lst, cv) { lmt.emplace_back(lst.size() / 2); }
	V<ll> ans_l;
	REPC(c, str) {
		ll it = c - static_cast<ll>('a');
		if (lmt[it] > 0) {
			ans_l.emplace_back(it);
			--lmt[it];
		}
	}

	ll ans = 0;
	REP(i, ans_l.size()) {
		ll c = ans_l[i];
		ll l = *cv[c].begin();
		ll r = *cv[c].rbegin();
		cv[c].pop_back();
		cv[c].pop_front();

		ans += l - i + tree.query(l, l);
		dump(ans);
		ans += (size - 1 - i - r - tree.query(r, r));
		dump(ans);
		tree.add(0, l, 1);
		tree.add(r, size, -1);
	}
	cout << ans << endl;
}