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-----
eel

-----Sample Output-----
1

We can turn S into a palindrome by the following operation:
 - Swap the 2-nd and 3-rd characters. S is now ele.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = '"""\nまず、奇数個存在する文字種が複数ある場合は、回文にできない。\nそれ以外の場合は、回文にできる。\n次に、操作の最小回数は、回文になった場合の各文字のインデックスを求めておき、BITを使って、入れ替えが必要な回数を求めて置く\n"""\nfrom collections import Counter\nS = input()\nN = len(S)\n#各文字の出現回数をカウント\napperance = Counter(S)\n\n#回文を作成できるか判定\nflag = False\nfor k,v in list(apperance.items()):\n    if v%2==1:\n        if flag:\n            print((-1))\n            return\n        else:\n            flag = True\n            t = k\n\n\n#place[s] -> 文字sの出現位置\nplace = {}\nfor i in range(N):\n    s = S[i]\n    if s not in place:\n        place[s] = []\n    place[s].append(i)\n\n#memo[i] -> Sのi文字目の回文上でのインデックス\nmemo = [-1] *(N)\n\n#回文前半、後半に含まれることになる文字に、回文上でのインデックスを割り当てる。\ntmp = 0\ntmpApperance = {}\nfor i in range(N):\n    s = S[i]\n    if s not in tmpApperance:\n        tmpApperance[s] = 1\n    else:\n        tmpApperance[s] += 1\n    if tmpApperance[s] <= apperance[s]//2:\n        memo[i] = tmp\n        backIdx = place[s][-tmpApperance[s]]\n        memo[backIdx] = N-tmp-1\n        tmp += 1\n\n\n#回文の真ん中に含まれることになる文字に、回文上でのインデックスを割り当てる。\nif flag:\n    idx = N//2\n    memo[place[t][len(place[t])//2]] = idx\n\nclass BIT:\n    def __init__(self,n):\n        self.size=n\n        self.tree = [0]*(n+1)\n    \n    def sum(self,i):\n        s = 0\n        while i>0:\n            s += self.tree[i]\n            i -= i&-i\n        return s\n    \n    def add(self,i,x):\n        while i<=self.size:\n            self.tree[i] += x\n            i += i&-i\nbit = BIT(N)\ncnt = 0\nfor i in range(N):\n    m = memo[i]+1\n    bit.add(m,1)\n    cnt += bit.sum(N)-bit.sum(m)\n\nprint(cnt)\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: