description: Petya has got an interesting flower. Petya is a busy person, so he sometimes forgets to water it. You are given n days from Petya's live and you have to determine what happened with his flower in the end.
  
  The flower grows as follows: 
  
    * If the flower isn't watered for two days in a row, it dies. 
    * If the flower is watered in the i-th day, it grows by 1 centimeter. 
    * If the flower is watered in the i-th and in the (i-1)-th day (i > 1), then it grows by 5 centimeters instead of 1. 
    * If the flower is not watered in the i-th day, it does not grow. 
  
  
  
  At the beginning of the 1-st day the flower is 1 centimeter tall. What is its height after n days?
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
  
  The first line of each test case contains the only integer n (1 ≤ n ≤ 100).
  
  The second line of each test case contains n integers a_1, a_2, ..., a_n (a_i = 0 or a_i = 1). If a_i = 1, the flower is watered in the i-th day, otherwise it is not watered.
  
  Output
  
  For each test case print a single integer k — the flower's height after n days, or -1, if the flower dies.
  
  Example
  
  Input
  
  
  4
  3
  1 0 1
  3
  0 1 1
  4
  1 0 0 1
  1
  0
  
  
  Output
  
  
  3
  7
  -1
  1
solution: import sys,os,io
  input = sys.stdin.readline # for strings
  # input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline # for non-strings
  
  PI = 3.141592653589793238460
  INF =  float('inf')
  MOD  = 1000000007
  # MOD = 998244353
  
  
  def bin32(num):
      return '{0:032b}'.format(num)
  
  def add(x,y):
      return (x+y)%MOD
  
  def sub(x,y):
      return (x-y+MOD)%MOD
  
  def mul(x,y):
      return (x*y)%MOD
  
  def gcd(x,y):
      if y == 0:
          return x
      return gcd(y,x%y)
  
  def lcm(x,y):
      return (x*y)//gcd(x,y)
  
  def power(x,y):
      res = 1
      x%=MOD
      while y!=0:
          if y&1 :
              res = mul(res,x)
          y>>=1
          x = mul(x,x)
          
      return res
          
  def mod_inv(n):
      return power(n,MOD-2)
  
  def prob(p,q):
      return mul(p,power(q,MOD-2))    
    
  def ii():
      return int(input())
  
  def li():
      return [int(i) for i in input().split()]
  
  def ls():
      return [i for i in input().split()]
  
  for t in range(ii()):
      t+=1
      n = ii()
      a = li()
      h = 1
      for i in range(n):
          if a[i] == 1:
              if i != 0:
                  if a[i-1] == 1:
                      h+=5
                  else:
                      h+=1
              else:
                  h+=1
          else:
              if i != 0 and a[i-1] == 0:
                  h=-1
                  break
      print(h)