A binary string is called a self-destructing string if it can reduced to an empty string by performing the following operation some number of times (possibly zero): Choose a valid integer $i$ such that the $i$-th character of the current string is different from the $i+1$-th character, and remove these two characters from the string.
You are given a binary string $s$. Your task is to convert $s$ to a self-destructing string. To do that, you may perform the following operation any number of times (possibly zero): Choose an integer $i$ ($1 \le i \le |s|-1$) such that the $i$-th character of $s$ is different from the $i+1$-th character, and invert one of these characters (inverting a character means changing '0' to '1' or '1' to '0', e.g. the string "01" can be changed to "00").
Find the smallest number of operations required to convert $s$ to a self-destructing string or determine that it is impossible.

-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first and only line of each test case contains a single string $s$.

-----Output-----
For each test case, print a single line containing one integer ― the smallest required number of operations or $-1$ if it is impossible to convert $s$ to a self-destructing string.

-----Constraints-----
- $1 \le T \le 1,000$
- $1 \le |s| \le 10^6$
- $s$ contains only characters '0' and '1'
- the sum of $|s|$ over all test cases does not exceed $10^6$

-----Example Input-----
3
001101
1101
110

-----Example Output-----
0
1
-1

-----Explanation-----
Example case 1: We already have a self-destructing string because "001101" can be converted to "0101", to "01" and finally to an empty string.
Example case 2: We can choose $i=3$ (since $s_3 \neq s_4$) and then invert $s_4$ to obtain "1100", which is a self-destructing string.
Example case 3: It can be shown that "110" cannot be converted to a self-destructing string.

```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 = "# 10 + 20 + 2:28\n\n\n\ndef solve(b,n):\n x=0\n y=0\n for i in range(n):\n  if b[i]=='1':\n   x+=1\n  else:\n   y+=1\n if x==n or y==n:\n  return -1\n\n return abs(x-y)//2\n\n\nfor tests in range(int(input())):\n b=input()\n n=len(b)\n if n%2==1:\n  print(-1)\n else:\n  print(solve(b,n))"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13077