description: <image>
  
  William has an array of n integers a_1, a_2, ..., a_n. In one move he can swap two neighboring items. Two items a_i and a_j are considered neighboring if the condition |i - j| = 1 is satisfied.
  
  William wants you to calculate the minimal number of swaps he would need to perform to make it so that the array does not contain two neighboring items with the same parity.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
  
  The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) which is the total number of items in William's array.
  
  The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) which are William's array.
  
  It is guaranteed that the sum of n over all test cases does not exceed 10^5.
  
  Output
  
  For each test case output the minimal number of operations needed or -1 if it is impossible to get the array to a state when no neighboring numbers have the same parity.
  
  Example
  
  Input
  
  
  5
  3
  6 6 1
  1
  9
  6
  1 1 1 2 2 2
  2
  8 6
  6
  6 2 3 4 5 1
  
  
  Output
  
  
  1
  0
  3
  -1
  2
  
  Note
  
  In the first test case the following sequence of operations would satisfy the requirements: 
  
    1. swap(2, 3). Array after performing the operation: [6, 1, 6] 
  
  
  
  In the second test case the array initially does not contain two neighboring items of the same parity.
  
  In the third test case the following sequence of operations would satisfy the requirements: 
  
    1. swap(3, 4). Array after performing the operation: [1, 1, 2, 1, 2, 2] 
    2. swap(2, 3). Array after performing the operation: [1, 2, 1, 1, 2, 2] 
    3. swap(4, 5). Array after performing the operation: [1, 2, 1, 2, 1, 2] 
  
  
  
  In the fourth test case it is impossible to satisfy the requirements.
  
  In the fifth test case the following sequence of operations would satisfy the requirements: 
  
    1. swap(2, 3). Array after performing the operation: [6, 3, 2, 4, 5, 1] 
    2. swap(4, 5). Array after performing the operation: [6, 3, 2, 5, 4, 1] 
solution: from __future__ import division, print_function
  import threading,bisect,math,heapq,sys,random
  # threading.stack_size(2**27)
  # sys.setrecursionlimit(10**4)
  from sys import stdin, stdout
  i_m=9223372036854775807    
  def cin():
      return map(int,sin().split())
  def ain():                           #takes array as input
      return list(map(int,sin().split()))
  def sin():
      return input()
  def inin():
      return int(input()) 
  
  
  """**************************MAIN*****************************"""
  def main():
      t=inin()
      for _ in range(t):
          n=inin()
          a=ain()
          b=[]
          for i in range(n):
              a[i]=a[i]%2
              if(a[i]!=0):
                  b.append(i)
          c=b[::-1]
          ans=0
          for i in range(1,n,2):
              if(len(c)==0):
                  ans=i_m
                  break
              ans+=abs(c.pop()-i)
          if len(c)>0:
              ans=i_m
          c=b[::-1]
          an=0
          for i in range(0,n,2):
              if(len(c)==0):
                  an=i_m
                  break
              an+=abs(c.pop()-i)
          if len(c)>0:
              an=i_m
          ans=min(ans,an)
          if ans==i_m:
              ans=-1
          print(ans)
          
  """*******************************************************"""
  
  ######## Python 2 and 3 footer by Pajenegod and c1729
  fac=[]
  def fact(n,mod):
      global fac
      fac.append(1)
      for i in range(1,n+1):
          fac.append((fac[i-1]*i)%mod)
      f=fac[:]
      return f
  def nCr(n,r,mod):
      global fac
      x=fac[n]
      y=fac[n-r]
      z=fac[r]
      x=moddiv(x,y,mod)
      return moddiv(x,z,mod)
  def moddiv(m,n,p):
      x=pow(n,p-2,p)
      return (m*x)%p
  def GCD(x, y): 
      x=abs(x)
      y=abs(y)
      if(min(x,y)==0):
          return max(x,y)
      while(y): 
          x, y = y, x % y 
      return x 
  def Divisors(n) : 
      l = []  
      ll=[]
      for i in range(1, int(math.sqrt(n) + 1)) :
          if (n % i == 0) : 
              if (n // i == i) : 
                  l.append(i) 
              else : 
                  l.append(i)
                  ll.append(n//i)
      l.extend(ll[::-1])
      return l
  def SieveOfEratosthenes(n): 
      global prime
      prime = [True for i in range(n+1)] 
      p = 2
      while (p * p <= n): 
          if (prime[p] == True): 
              for i in range(p * p, n+1, p): 
                  prime[i] = False
          p += 1
      f=[]
      for p in range(2, n): 
          if prime[p]: 
              f.append(p)
      return f
  def primeFactors(n): 
      a=[]
      while n % 2 == 0: 
          a.append(2) 
          n = n // 2 
      for i in range(3,int(math.sqrt(n))+1,2):  
          while n % i== 0: 
              a.append(i) 
              n = n // i  
      if n > 2: 
          a.append(n)
      return a
  """*******************************************************"""
  py2 = round(0.5)
  if py2:
      from future_builtins import ascii, filter, hex, map, oct, zip
      range = xrange
  import os
  from io import IOBase, BytesIO
  BUFSIZE = 8192
  class FastIO(BytesIO):
      newlines = 0
   
      def __init__(self, file):
          self._file = file
          self._fd = file.fileno()
          self.writable = "x" in file.mode or "w" in file.mode
          self.write = super(FastIO, self).write if self.writable else None
   
      def _fill(self):
          s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
          self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
          return s
   
      def read(self):
          while self._fill(): pass
          return super(FastIO,self).read()
   
      def readline(self):
          while self.newlines == 0:
              s = self._fill(); self.newlines = s.count(b"\n") + (not s)
          self.newlines -= 1
          return super(FastIO, self).readline()
   
      def flush(self):
          if self.writable:
              os.write(self._fd, self.getvalue())
              self.truncate(0), self.seek(0)
   
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          if py2:
              self.write = self.buffer.write
              self.read = self.buffer.read
              self.readline = self.buffer.readline
          else:
              self.write = lambda s:self.buffer.write(s.encode('ascii'))
              self.read = lambda:self.buffer.read().decode('ascii')
              self.readline = lambda:self.buffer.readline().decode('ascii')
   
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip('\r\n') 
  # Cout implemented in Python
  class ostream:
      def __lshift__(self,a):
          sys.stdout.write(str(a))
          return self
  cout = ostream()
  endl = '\n'
   
  # Read all remaining integers in stdin, type is given by optional argument, this is fast
  def readnumbers(zero = 0):
      conv = ord if py2 else lambda x:x
      A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
      try:
          while True:
              if s[i] >= b'R' [0]:
                  numb = 10 * numb + conv(s[i]) - 48
              elif s[i] == b'-' [0]: sign = -1
              elif s[i] != b'\r' [0]:
                  A.append(sign*numb)
                  numb = zero; sign = 1
              i += 1
      except:pass
      if s and s[-1] >= b'R' [0]:
          A.append(sign*numb)
      return A
   
  # threading.Thread(target=main).start()
  if __name__== "__main__":
    main()
  